PLANE AND SPHERICAL 
TRIGONOMETRY 



BY 

A. H. BUCHANAN., LL.D. 

PROFESSOR OF MATHEMATICS 
CUMBERLAND UNIVERSITY 



FIRST EDITION 

FIRST THOUSAND 



NEW YORK 

JOHN WILEY & SONS 

London : CHAPMAN & HALL, Limited 

1907 



*fc 



t* 






{ubSSyoTconsbi 

Two Copies Received 

DEC 10 *90? 

Couyngnt tntry 



Copyright, 1907, 

BY 

A. H. BUCHANAN 



Stanbope flftess 

F, H. G1LSON COMPAHT 
BOSTON. U.S. A 



PREFACE 



This work comprises all that is required in the curriculum for 
the A.B. degree, excepting Trigonometric Series and Hyperbolic 
Functions, which are given later in Calculus and properly belong 
there. 

Many of the demonstrations of principles and the determina- 
tions of formulas are new, whether more simple or not others 
must judge. Simplicity in no case has been sought at the expense 
of accuracy, but the most scientific accuracy is never appreciated 
by the student and should not be required of him. One demon- 
stration, that the most simple, and but one well understood, 
should limit the requirement of every teacher. If others are 
given, they should be as exercises. 

When the functions are correctly defined, the demonstrations 
for acute and obtuse values are the same, and therefore but one 
is necessary. 

Suggestions are given in the difficult exercises and examples, 
which it is believed the teacher will appreciate, as he must give 
these generally before the student can do anything. 

Examples have been selected from the various text-books: 
Chauvenet, Newcomb, Crawley, Olney, Went worth, Wells, Miller, 
Crockett, Bowser, Lock, etc. The examples have been solved 
with five-place tables, so the results will differ slightly from those 
obtained by the larger tables. 

The text is bound without the tables, as the student must use 
these in his daily tests and final examinations; and, besides, a 
separate binding is more convenient in other parts of his course. 

A. H. Buchanan. 

June, 1907. 



TABLE OF CONTENTS 



INTRODUCTION 

Page. 

Linear and angular units 1 

Generation of angles and arcs 2 

Lines, when plus; when minus 3 

Examples 3 

CHAPTER I 

Angles in the four quadrants 5 

Trigonometric ratios defined 6 

Signs of the functions in the four quadrants 7 

Reciprocal relations of the functions 8 

Functions of an angle in the four quadrants 9 

Functions represented by lines 11 

Values and signs of the functions in the four quadrants 12 

Functions of 45° and 60° 13 

Formulae grouped for reference 14 

Exercises 17 

CHAPTER II 

Solutions of right triangles 20 

Areas of right triangles 23 

Examples and exercises 23 

CHAPTER III 

Determinations of fundamental formulae 26 

Functions of two angles 29 

Functions of twice an angle 30 

Functions of half an angle 31 

Functions of three and four times an angle 32 

Exercises 33 

Inverse trigonometric functions 36 

v 



VI TABLE OF CONTENTS 

CHAPTER IV 

Page. 

Formulae for the solution of oblique triangles 38 

Methods of solving oblique triangles 43 

Areas of oblique triangles 48 

Examples in oblique triangles and exercises 49 

CHAPTER V 

Circumscribed and inscribed circles — triangle and quadrilateral . 54 

Solutions of trigonometric equations 56 

Computations of trigonometric functions 57 

De Moivre's Theorem 58 

CHAPTER VI 

Spherical triangles defined 60 

Geometry of the spherical triangle 61 

Formulae for the spherical right triangle 63 

Napier's rules 65 

Limitations between the parts of right spherical triangles .... 66 

Quadrantal triangles 68 

Examples of spherical right triangles 68 

CHAPTER VII 

Relations of the parts of the general spherical triangle 74 

General formulae adapted to logarithms 77 

Napier's Analogies 81 

Gauss's Equations 81 

CHAPTER VIII 

Limitations between the parts of the general spherical triangle . . 82 

Solutions of the oblique spherical triangle 83 

Ambiguous case of Case III 85 

Ambiguous case of Case IV 88 

Area of the general spherical triangle 91 

Circumscribed and inscribed circles of spherical triangles 92 

Examples and exercises 94 



PART I 

PLANE TRIGONOMETRY 

INTRODUCTION 

1. Trigonometry treats of the functions of angles, and sub- 
jects them to numerical computation in the solution of triangles. 

Plane Trigonometry treats of plane angles and the solution of 
plane triangles. 

2. The sides of triangles are measured in any of the linear 
units, as feet, meters, yards, rods, etc. 

3. Angles are usually measured in the sexagesimal units, — 
degrees, minutes, and seconds. 

One degree = -^ of a right angle, 
One minute = F V of one degree, 
One second = ^V °f one minute. 

4. The centesimal method, which is seldom used, divides the 
right angle into 100 equal parts called grades; a grade is divided 
into 100 minutes; and a minute into 100 seconds. 

Example. 73 g 86' 95" = 73 f 8695. 

5. The circular method, used in higher 

mathematics, is expressed by arc . 

radius 
In this, angles are measured by the arcs of 
circles which subtend them; the unit called 
the radian being an arc whose length is 
the same as that of its radius. Therefore 
180° is n or 3.1416 radians, 90 is \ tz 
radians, and 360 is 2 tz radians. The value 
cf one radian is — = 57° 17' 44.8" units of the sexagesimal 

TZ 

method. Hence the rule, 




PLANE TRIGONOMETRY 



To convert degrees into radians, 



convert radians into degrees, multiply by 



multiply by 
180° 



180° 



and to 



6. If a straight line OB, starting from coincidence with OA, 
revolves in the same plane about the point 0, Fig. 2, it is said to 
generate in succession the angles AOB, AOC, AOD, etc., the meas- 
ures of which may be expressed in degrees, minutes, and seconds. 
Or the arcs A'P, A'P', A'P", A'P"> ', A'P 1 ", etc., generated in this 

revolution by a point P at a 
unit's distance from 0, may 
be taken as the measures of 
the same angles expressed 
in radians. 

Angles or arcs may be 
generated by turning the 
line OB in the opposite 
direction, in which case 
the angles AOH, AOG, 
AOF, etc., and the arcs 
A'P™, A'P", A'P™, etc., 
must be considered nega- 
tive. In other words, 
angles or arcs estimated in 
the direction contrary to 
the motion of the hands of 
a clock, or counter clockwise, are called positive, while those 
estimated in the direction of their motion, or clockwise, are 
called negative. 

The possible rotation of the line OB or of its point P being 
without limit, the magnitude of the angle or arc generated is not 
limited; that is, an angle may be one, two, three, four, or any 
number of right angles, and an arc may be \ n, n, f n, 2 n, or any 
number of times n. 

AR is called the initial diameter, and the point A', the origin 
of arcs. 

7. If a line drawn in a given direction is reckoned positive, 
when drawn in the opposite direction it must be reckoned nega- 




PLANE TRIGONOMETRY 



3 



p' 



live. In Fig. 3, the two indefinite lines XX' and YY', called 
coordinate axes, are drawn at right angles to each other. Dis- 
tances measured to the right of YY' and parallel to XX' are 
positive, while those to the left and parallel to XX' are negative. 
Distances measured upward and parallel to YY' are positive; 
those downward and parallel to YY' are negative, both being 
measured from XX'. Hence OE, OA, BP, are positive; OC, OD, 
FP", are negative. AP, OB, DP', are positive; CP", OF, EP'", 
are negative. If OB, equal 
AP, is a inches long, either y 

may be represented by + a; 
and if OF, equal to CP", is 
b inches long, either may be 

represented by — b. Likewise x- 

OA equal to BP may be 
denoted by + c, and OC equal 
to FP" by - d, etc. 

8. The complement of an 
angle or an arc is the remain- 
der after subtracting it from 
90° or ^ 7i, and may be posi- 
tive or negative, according as it is less or greater than 90° or \ tz. 

The supplement of an angle or arc is the remainder after 
subtracting it from 180° or n, and may also be positive or 
negative. 

For example, the complement of 30° is 60°; of 150° is - 60°. 
The supplement of 150° is 30°; of 250° is - 70°. 



Fig. 3 



EXAMPLES 

1. What are the complements of 37°, 17°, 36°, 45°, 29°, 27°, 32°, 216°, 
97°? 

2. What are the supplements of 128°, 37° 4' 3", 115° 13' 25", 226° 14' 
17", 189° 45' 37"? 

3. How many degrees, minutes, and seconds in it, \ tz, f tz, J tz, \ tz, \ re? 

4. How many radians in 18°, 90°, 120°, 1800°, 96°, 360°, 300°, 87° 27' 
57", 57° 17' 44. 8"? 

5. Find the number of degrees in the angle subtended by an arc 46 
feet and 9 inches long, at the center of a circle whose radius is 25 feet. 



4 PLANE TRIGONOMETRY 

6. Find the number of degrees in an angle at the center of a circle of 
25 feet radius, which stands on an arc of 37.5 feet. 

7. A railway train is traveling on a curve of half a mile radius at the 
rate of 20 miles an hour. Through what angle does it turn in 10 seconds? 

Ans. 6° 22'. 

8. One angle of a triangle is 45°, and the circular measure of another 

is 1.5 radians. Find the third angle. 

Ans. 49° 3' 23". 

9. The angles of a triangle are in arithmetic progression, and the 
greatest is double the least ; express the angles in degrees. 



CHAPTER I 
TRIGONOMETRIC RATIOS OR FUNCTIONS 

9. A quantity is a function of another when so connected with 
it as to derive its value from it. 

Draw the axes XX' and 77', Fig. 4, at right angles. Revolve 
the line OA once about 0, starting from coincidence with OX, 
and it will generate every possible angle between 0° and 360°. 
Any point P in it, starting from coincidence with B, will describe 




x D 



Fig. 4 



every possible arc from zero to the full circumference BPCP'P"P'". 
The angles made by these several positions of the line OA with 
OX are reckoned counter-clockwise from the initial OX. 

BOC, COD, DOE, and EOB are called respectively the I, II, 
III, and IV quadrants, and the angles corresponding to the 
several positions of the revolving line, as OA, OA f , OA" , and 
OA"' in these, are called angles in the I, II, III, or IV quadrants, 
according to the number in which its terminal side stands. Thus 

5 



6 PLANE TRIGONOMETRY 

XOA = in the I, XOA' = 0' in the II, XOA" = 0" in the 
III, and XOA'" = 0"' in the IV quadrant, each being indicated 
by its dotted curve in the above figure. 

10. Draw PR, P'R', P"R" , and P'"R'" perpendicular to XX', 
forming four right triangles OPR, OP'R', OP"R" ', and OP'"R'". 

From the hypotenuse, perpendicular and base of any one of 
these, or any right triangle whatever, six different ratios can be 
formed; three independent, thus: 

-I , perpendicula r 
hypotenuse ' 

2d. base 



hypotenuse ' 

Q , perpendicular 

oa. , 

base 

and three dependent, obtained by inverting these three; 
4th hypotenuse 
perpendicular ' 

5th hypotenuse 
base 

6th. base 



perpendicular 



These all being the ratios of one line divided by another are 
abstract numbers, and have received specific names as functions 
of the angle between the hypotenuse and base. 

Trigonometric Functions Defined 

If from any point in either side of an angle a perpendicular be 
drawn to the other side, or the other side produced, forming a 
right triangle, as POR in Fig. 4, then 

1st, Perpendicular ig caUed the sine Q j ^ and written sin q. 
hypotenuse 

2d, ^^ , is called the cosine of 6, and written cos 0; 

hypotenuse 

3d, perpendicular , is called the tangent of 6, and written tan 0; 
base 



TRIGONOMETRIC RATIOS OR FUNCTIONS 7 

4th, — *-*- — , is called the cosecant of d, and written esc 0; 

perpendicular 

5th, ypo enuse ^ - g ca ^ e( j ^ e secan i j n anc j written sec 6; 
base 

6th, — , is called the cotangent of 6, and written cot 6. 

perpendicular 

These ratios are called functions of 6 because their values 
depend upon its magnitude. 

11. The above definitions being general are alike applicable 
to angles in the several quadrants, Fig. 4, their algebraic signs 
being determined by the signs of the base and altitude of the 
triangles in each, their hypotenuses being always positive. 

In the triangle of the I quadrant, all the sides are positive, and 
the ratios are, 

a RP a OR , Q RP 

sin a = — — - , cos o = — — , tan o = — — , 
OP 1 OP 7 OR' 

a OP a OP , n OR 

™e=—, sectf=— , cot0= — , 

all positive. 

In the triangle of the II quadrant, OR' being negative, the 
ratios are, 

• m R'P' to -OR' . at R'P' 

sm 0' = — — , cos 6' = ^ - , tan 6' = 



OP' ' OP' ' - OR' 

csc 0'=-^, sec 6" = — ??1 cottf' = ~ 



R'P' ' -OR' ' R'P'' 

only sine and cosecant positive. 

In the triangle of the III quadrant, OR" and R"P" being 
negative, the ratios are, 

B infl"= *„„ , cos 6" = -^-, tan 0" = K F , 
OP" OP" -OR" ' 

csc 6"= 2 P " sec 0"=-2^-, cot 6" 



-R"P"' -OR" -R"P 

only tangent and cotangent positive. 



3 



PLANE TRIGONOMETRY 



In the triangle of the IV quadrant, R'"P'" being negative, the 
ratios are, 

-R'"P'" a „, OR'" 



sin 



esc 



OP'" ; 
OP"' 



cos 0'" = 



OP' 



, tan 



R'"P' 



sec 



OP'" 
ORT' COt 



R"'P'" 
OR'" ' 

OR"' 
R'"P'" ' 



only cosine and secant positive. 
These results tabulated are 



Quadrant. 


i. 


ii. 


in. 


IV. 


Sine and Cosecant 


+ 
+ 
+ 


+ 


+ 




Cosine and Secant 


+ 


Tangent and Cotangent • 





12. There are two other trigonometric ratios in use, called 
versed-sine and cover sed-sine, which are, Fig. 4, 



Versed-sine 



Co versed-sine d = 
These are written: 



RB _ 0B -OR 
OP OP 

CF PC -OF 
OP OP 



= 1 - 



OR 
OP 
OF 
OP 



1 



cos 6, 
sin d 



vers 0=1— cos d, 
covers 6 = 1 — sin 6. 



13. From the definitions, § 10, and the equations of § 11, the 
following mutual relations of the functions of any angle are 
already made evident, viz., 



sin o = 



CSC 

1 



or esc 6 = - — - , 
sin 



cos 6 = or sec d 

sec 



cos 
1 



Or, 



tan 6 = or cot d = 

cot o tan a 

Sine and Cosecant are reciprocal, 
Cosine and Secant are reciprocal, 
Tangent and Cotangent are reciprocal. 



TRIGONOMETRIC RATIOS OR FUNCTIONS 



14. Describe a circle, Fig. 5, with any radius; draw the axes 
XX' and YY' at right angles, and the lines A A" and A' A'" 
making the same acute angle of any value whatever, above and 
below XX'. Then XOA = 6, XOA' = 180° - 6, XOA" = 180° 
+ 0, and XOA'" = 360° - 6, each of which may be regarded 
as generated by the revolution of OA in the positive direction 
from OX. AA'" and A' A" are perpendicular to XX' and form 
the four right triangles AOC, A'OD, A"OD, and A'"OC. There- 
fore the functions of the angles 0, 
180° - 0, 180° + 0, and 360° - d, to 
which these triangles correspond, can 
differ only in their algebraic signs. 
Hence, as is seen in the figure, 



CA 

OA 



DA' 

OA' 



DA 



CA' 



OA- 



OA' 



in absolute value, and so for the other 
ratios which are left for the student to 
determine; or, 




sin = sin (180° - 0) = - sin (180° + d)= - sin (360° - 0;, 
cos = - cos (180° - 0) = - cos (180° + 0) = cos (360° - 0), 
tan = - tan (180° - 0) = tan (180° + 6) = - tan (360° - d), 
cot d = - cot (180° - 0) = cot (180° + 0) = - cot (360° - 6). 



If any multiple of 360° or 2 n be added to each of these angles, 
the same values of the functions corresponding to the first revo- 
lution of OA will apply to those of a second, third, or any num- 
ber, since they will always terminate at the same point in the 
same quadrant. 

The first and second terms of the above equations prove that, 

The sine of an angle = the sine of its supplement. 

The cosine of an angle = minus the cosine of its supplement. 

The tangent of an angle = minus the tangent of its supplement. 

The cotangent of an angle = minus the cotangent of its supple- 
ment, 

15. Draw AB in Fig. 5 perpendicular to YY' to form the right 
triangle BAO, equal in every respect to AOC. Let the angle 



10 PLANE TBIGONOMETRY 

AOY = 0'. Then = 90 — 0', and may therefore replace in 
the equations of the preceding paragraph, giving, 

sin (90 - 0') = -^f- = cos 6' = sin (90° + 0') 

= -sin (270° - 00 =- sin (270° + 00, 

cos (90 - 00 = T^r = sin 0' = - cos (90° + 00 

= - cos (270° - 00 = cos (270° + 00, 

tan (90 - 00 = 4rr = cot 6 ' = ~ tan ( 90 ° + ^ 

= tan (270° - 00 = ' - tan (270° + 00, 

cot (90 - 00 = -^4" = tan & = ~ cot ( 90 ° + e ') 
OB 

= cot (270° - 00 = - cot (270° + 0*). 

These being entirely general for any acute value of 0', its dash 
may therefore be left off. Then after changing places of the 
first and second equations and also of the third and fourth, 
interchanging the first and third terms in all and omitting the 
second terms, the following group results: 

sin = cos (90° - 0) = - cos (90° + 0) 

= - cos (270° - 0) = cos (270° + 0),. 
cos = sin (90° - 0) = sin (90° + 0) 

= - sin (270° - 0) = - sin (270° + 0), 
tan = cot (90° - 0) = - cot (90° + 0) 

= cot (270° - 0) = - cot (270° + 0), 
cot = tan (90° - 0) = - tan (90° + 0) 

= tan (270° - 0) = - tan (270° + 0). 

The remark as to multiples of 360° or 2 n applies to these as 
in § 14. By the first and second columns in these equations it 
appears that, 

The sine of an angle = the cosine of its complement. 
The cosine of an angle = the sine of its complement. 
The tangent of an angle = the cotangent of its complement. 
The cotangent of an angle = the tangent of its complement. 



TRIGONOMETRIC RATIOS OR FUNCTIONS 



11 



If the line OA, Fig. 5, be turned in the negative direction to 
coincide with OA" r , it generates a negative angle (— 6), whose 
functions are evidently the same as those of 360° — 6, § 14. 



Hence 

sin (—6) = — sin 6, and therefore esc ( — 0) = — esc 6, 
cos ( — 6) — cos 6, and therefore sec ( — 0) == sec 6, 
tan ( — 0) = — tan 6, and therefore cot ( — 0) = — cot d. 

Corollary. If in Fig. 5 the radius 
OA be taken one unit in length, all 
the triogonometric ratios may be 
represented by lines about the 
quadrant in which the angle is 
found. Draw the quadrant, Fig. 6, 
with a radius of one inch, and 
complete the figure. 

The triangles OAC, OBD, and OEF 
are similar. Then since OB, OA, 
and OF are each = 1, and AOB = 0, 




sin 



CA 
OA 



= CA, 



tanfl = ^ = BD, 
seed =^=OD, 
vers 6 = ^ = CB, 

DA. 



cos 6 



cot = 



CSC u = 



covers u = 



OA~° C ' 

OF 

OF 



GF 
OA 



GF. 



16. If the right triangle AOC, Fig. 6, has sides of any length 
whatever, 

sin 6 — — — - , and cos 6 = — — - ; therefore 
OA OA 



sin CA 



OC 



cosfl OA OA 



CA 

OC 



BD 



OB 



= tan 0. 



Hence, 



, a sin 
tan = — — , or 
cos o 



12 



PLANE TRIGONOMETRY 



The tangent of any angle is equal to its sine divided by its cosine. 

From this and §13 it follows that all the other functions of any 
angle with their algebraic signs may be determined from its sine 
and cosine. 

Functions of 0°, 90°, 180°, and 270°. 

17. In the equations of § 14 if is made equal to zero, the 
bases of the triangles in Fig. 5 in passing from the fourth to the 
first quadrants, and from the second to the third quadrants, will 
become equal to the hypotenuses and their altitudes equal to 
zero. Then from it and the formula will result 

sin 0° = =F 0= ± sin 180°, and therefore esc 0° = =F oo = ± esc 180°, 
cos 0° = + 1 - - cos 180°, " " sec 0° = + 1 = - sec 180°, 

tan 0° = =F 0= =F tan 180, " " cot = =F oo = =F cot 180. 

Likewise if be made zero in the equations of § 15 the results 
will be, 

sin 0° = =F = ± cos 90° = =f cos 270°, and 

esc 0° = =F oo = ± sec 90° = =F sec 270°, 

cos 0° = + 1 = + sin 90° = - sin 270°, and 

sec 0° = + 1 = + esc 90° = - esc 270°, 

tan 0° = =F = ± cot 90° = =F cot 270°, and 

cot 0° = =F oo = ± tan 90° = ± tan 270°. 

These results are tabulated for convenient reference. 





sine 


cosine 


tang. 


cotan. 


secant 


cosec. 


0° 


TO 


+ 1 


TO 


=f oo 


+ 1 


±00 


90° 


+ i 


±o 


±00 


±0 


=b'« 


+ 1 


180° 


±o 


- 1 


TO 


T*o 


- 1 


±oc 


270° 


- l 


=F0 


±oo 


±0 


T*> 


- 1 



The double sine before oo and means the function changes 
from plus to minus or from minus to plus in passing the given 
value of the angle. 

Example. Tan 90° = ± oo means the tangent is + oo just 
as the angle reaches 90°, but becomes — oo just as it is passing 90°. 



TRIGONOMETRIC RATIOS OR FUNCTIONS 



13 



Another table easily derived from the above shows the varia- 
tion in the values of the functions as the angle increases. 



Quadrants 


i 


ii 


in 


IV 




varies from 


varies from 


varies from 


varies from 


sine 


+ to + 1 


+ 1 to + 


- to - 1 


- 1 to - 


cosine . . 






+ 1 to + 


- to - 1 


- 1 to - 


+ to + 1 


tangent . 






+ to + oo 


- oo to - 


+ to + oo 


— oo to — 


cotangent 






+ oo to + 


- to - GO 


+ oo to + 


- to - oo 


secant . . 






+ 1 to + oo 


- oo to - 1 


— 1 to — 00 


+ oo to + 1 


cosecant . 






+ 00 to + 1 


+ 1 to + oc 


— GO tO - 1 


- 1 to - oo 


versed-sine 






+ to + 1 


+ 1 to + 2 


+ 2 to + 1 


+ 1 to + 


coversed-sin( 






+ 1 to + 


+ to + 1 


+ 1 to + 2 


+ 2 to + 1 



18. Let the right triangle, Fig. 7, represent any one of those 
in the several quadrants of the preceding figures; denote the 
hypotenuse by h, the base by b, and the altitude by a. Then 
from § 10, 

a a _ „ j Q b b ± _ _. a a 

h 



suit/ = 



and cos 



h vV +b 2 ll Va 2 + b 2 

and the second values above show how 
the sine and cosine may be obtained from 
the tangent when given in the form of 
a fraction. 

From h 2 = a 2 + b 2 , by division 1 = 

a 2 b 2 

-- + — , whence by substitution 

h 2 h 2 



tan 




sin' 



> and 



sin 



-VI 



cos- 



+ COS' 
COS 6 = \/l 



sin' 



From above equation — = — 



1, and— =— + 1, then since 



sec 



sec' 



h 
b 
1 + tan 2 



and esc = - , these substituted give 
a 

d and esc 2 6=1 + cot 2 0. 
Functions of 45°. 



19. The complement of 45° is 45°, therefore, § 15, 
sin 45° - cos 45°, and by § 18, 

1 

V2 



sin 45 c 



cos 45 c 



V27 

9 ' 



14 



PLANE TRIGONOMETRY 



tan 45° = ^-^° = 1 = cot 45°, 



sec 45° 



cos 45 

1 
cos 45° 



= V2 = esc 45. 



Functions of 30° and 60°. 



20. Construct an equilateral triangle ABC, Fig. 8, and draw 

AD perpendicular to the base AC, bise cting it at D. AD = J AC 

= i AB, and BD = VAC 2 - \ AB 2 = } Vs AB. By §§ 11, 
13, 15, 

sin 30° = 4^ = I = cos 60°, 
AB 2 

cos 30° = ^? = - V3 = sin 60°, 
AB 2 




sec 30° = 



esc 30° = 



1 



sin 30 c 1 

cos 30° V3 

cos 30° 
sin 30 

2 



= cot 60°, 



b = \/3 = tan 60°, 



cos 



30° V3 



= esc 60°, 



1 



sin 30° 



= 2 = esc 60°. 



21. 



10. 



sin 



Formula Grouped for Convenient Reference 
a _ perpendicular „ = hypotenuse 



cos d 



hypotenuse 

base 
hypotenuse ' 



sec 



perpendicular ' 
hypotenuse 
base 
base 



taaf) = perpendicular ^ g = _ 

base perpendicular 



§ 12. 



vers # = L — cos d, 



covers 



1 — sin 



:l 



(i) 



(2) 



TRIGONOMETRIC RATIOS OR FUNCTIONS 

1 



13. 



sin 



cos o = 



tan = 



1 

CSC ' 

1 

sec 0' 

1 
cot0' 



CSC (/ = 



sec 



cot = 



sin 

1 

cos 

1 

tan 



15 



(3) 



§§16,18. 



tan 



sin 
cos 



cot0 = 



cos 



sin 



tan = - , sin = — = , 
b vV + 6 2 



COS (7 = 



Va 2 + b 2 



(4) 



'sin = sin (180° - 0) = -sin (180° + 0) 

= - sin (360° - 0), 
cos = - cos (180° - 6) = - cos (180° + 0) 

= cos (360° - 0), 
tan = - tan (180° - 0) = tan (180° + 0) 

= - tan (360° - 0), 
cot = - cot (180° - 0) = cot (180° + 0) 

= - cot (360° - 0), 
sec = - sec (180° - 0) = - sec (180° + 0) 

= sec (360° - 0). 
esc = esc (180° - 0) = - esc (180° + 0) 

= - esc (360° - 0). 



(5) 



15, 13. < 



sin = cos (90° - 0) = * - cos (90° + 0) 

= - cos (270° - 0) = cos (270° + 0), 
cos = sin (90° - 0) = sin (90° + 0) 

= - sin (270° - 6) = - sin (270° + 0), 
tan = cot (90° - 0) = - cot (90° + 0) 

= cot (270° - 0) = - cot (270° + 0), 
cot = tan (90° - 0) = - tan (90° + 0) 

= tan (270° - 0) = - tan (270° + 0), 
sec = esc (90° - 0) = esc (90° + 0) 

= - esc (270° - 0) = - esc (270° + d), 
esc - sec (90° - 0) = - sec (90° + 0) 

- - sec (270° - 0) = sec (270° + 0). J 



►(6) 



16 



PLANE TRIGONOMETRY 



Remark. It is evident from (5) and (6) that the functions of 
6, and 6 subtracted from or added to an even multiple of 90°, 
have the same names, but with an odd multiple the names are 
changed, sine to cosine, cosine to sine, tangent to cotangent, 
cotangent to tangent, etc. 

Functions of 0°, 90°, 180°, 210°. 



§17. 






sine 


cosine 


tangent 


cotangent 


secant 


cosecant 





1 





00 


1 


00 


90 


1 





00 





00 


1 


180 





-1 





— 00 


-1 


00 


270 


-1 





00 





— 00 


-1 



(7) 




18. 



h 2 =a 2 + b\ 

sin 2 6 + cos 2 6=1, 

sin 6 = Vl — cos 2 



cos 



e =vi 



sim 



sec 2 0=1 + tan 2 6, 
[esc 2 6=1 + cot 2 6. 



(8) 



'sin (— 6) = — sin 6, esc (— 6) = — esc 6, 
15. -{cos (— 6) = cos 6, sec (— 6) = sec #, 

tan(- 6) = - tan 6, cot (- 6) = - cot 6. 



(9) 



19. 



sin 45° = cos 45° = - \/2 sin 30° = cos 60° = - , 

2 v ^' 2 

tan 45° = cot 45° = 1, cos 30° = sin 60° =- Vs, 

sec 45° = esc 45° = V2, tan 30° - cot 60° = —j I qqx 

cot 30° = tan 60° = V3 



sec 30° = esc 60°=—, 

esc 30° == sec 60° = 2. J 



TRIGONOMETRIC RATIOS OR FUNCTIONS 
Each Function in Terms of the Others 



17 



In Terms 
of = 


sine 


cosine 


tangent 


cotangent 


secant 


cosecant 








tan0 


1 

\/l+COt2 

cot 








\/sec2 0-1 
sec 

1 

sec 


1 


sine . . 


\/l - cos-' ° 


\/l-ftan2 


CSC 




\/cSC2 0-1 




\/l - sin2 
sin 




\/l+tan2 

1 
tan 

\/l-|-tan 2 


\/l+COt2 
1 


CSC 

1 


tangent 


\/l-COS2 
COS 

COS 


cot 


Vsec20-1 

1 


V/CSC2 0-1 (11) 


cotan. . 


\/l — sin2 fl 
sin 

1 




V 1 — cos ' 2 


\Zsec2 — 1 
sec 


\/csc 2 — 1 




V'l-f-cot 2 ^ 
cot 


CSC 




\/l-sin2 

1 
sin 


COS 

1 

VI — cos-' 


\/cSC2 0-1 




\/l + tan-' 
tan 




cosec . 


\/i+<;ot2 


\/sec2 — 1 





EXERCISES 

1. In what quadrant are 120°, 340°, 490°, - 75°, - 360°? 

2. What signs belong to the different functions of each of the angles, 

- 281°, 367°, 508°, 650°, - 70°, - 197°? 

3. In what quadrants must an angle be whose secant is — 3? Whose 
tangent is - 2? Whose secant is 3? 

4. Express each of these in functions of A, — sin (180 + A), cos 
(270 - A), esc (360 - A): 

5. Construct the angles, 75°, 100°, 120°, 200°, - 75°. 

6. Give the signs of the sine, cosine, and tangent of 135°, 265°, 275°, 

- 10°. 

7. Draw any right triangle, ABC, right-angled at C, measure its 
sides and the angle BAC as carefully as possible, and then prove numeri- 
cally that AC 2 + BC 2 = AB 2 . Also prove numerically by the table of 

jyri a ri . T>ri 

natural functions that sin A = -— ■ , cos A = —-, and tan A = —~ . 

AB AB £C 

8. The sides of a right triangle are in the ratio of 5 : 12 : 13. Find 
the sine, cosine, and tangent of its acute angles. 

9. Prove the sine of 60° is equal to the sine of 120°, and sin 340° 
= sin (- 160°). 



18 PLANE TRIGONOMETRY 

10. If A = 90°, B = 60°, C = 30°, and D = 45°, prove, 

(1) 2 sin D cos D = sin A, 

(2) cos 2 B - sin 2 B = 1 - 2 sin 2 5, 

(3) 2 sin C cos C = sin 5, 

(4) sin B cos C + cos B sin C = sin A. 

11. Prove (tan A 4- cot A) sin A cos A = 1. 

12. Prove cos 2 A - sin 2 A = 2 cos 2 A - 1 = 1 — 2 sin 2 A. 

13. Prove cos 4 A - sin 4 J. = 2 cos 2 A — 1. 

14. Find the other functions if sin = 1, 
cos = VI - sin 2 = \/l - is =f. 

tan (9 = = — -4- — =— . For others, see §13. 

cos 5 5 4 

15. Find the other functions if cos = \. 

16. Find the other functions if sec = 4. _See (11). 

17. Find the other functions if tan 6 = Vs. See (11). 

18. Find the other functions if sin 6 = — . See (11). 

c 

19. Given tan A = cot A, find A. 

20. Find the numerical values of the sines, cosines, and tangents of 
150°, 135°,- 240°, 330°. 

21. Draw an angle whose sine is ^. 

22. Draw an angle whose tangent is 2. 

23. Prove the sin 70° = cos 20°. 

24. Prove the tan 79° = cot 11°. 

25. Find the angle which satisfies 2 cos 6 = sec 6. 

26. Find the angle which satisfies 3 sin 6 =2 cos 2 6. 

27. Find the angle which satisfies 4 sin 2 6 + 2 sin 6 = 1. 

28. Given: 8 sin 2 d - 10 sin 6 + 3 = 0, find sin d. 

29. In a right triangle ABC, find b and h if cot A = 4, and a = 17. (8) 

30. Find a and & if sec A ■= 2 and 6 = 20. 

31. Find sin d and cos 6, Fig. 7> if a + b = f h. (Divide by h and 
subst. (1). 

32. Find in degrees from esc d - cot 2 + 1=0. (9) 

33. Find the functions of d from tan + 3 cot d = 4. 

34. Find the functions of d from cos 2 d — \Z% cos 8 + f = 0. 

35. Find the functions of 6 from 4 sin 2 6 + 2 sin 6 = 1. 

36. Find the functions of from 2 sin 2 + ^2 sin 6 = 2. 

37. In the triangle, Fig. 7, if sin 6 = f and h = 20.5, find a and 6. 

38. Sin 2 = - i, find four values of 6 less than 360°. 

39. Tan 3 = 1, find six values of less than 360°. 



TRIGONOMETRIC RATIOS OR FUNCTIONS 19 

40. Find the value of , when vers x = - and x is in the 

cos a; X esc a; 4 

fourth quadrant. 

ai v j iL i £ sec x + sin x 

41. Find the value of — — 

1 - cot X 

third quadrant. 

42. Prove sin 3 + cos 3 = (sin + 

43. Prove cos 3 - sin 3 = (cos - 







Ans. + 


15. 


when tan 


X = 


2, and a; 


is in 


the 






Ans. 


_ li 
5 


V5. 


cos 0) (1 - 


sin 


0COS0). 






sin 6) (1 + sin 


L cos d). 







CHAPTER II 



23, 

§21, 



SOLUTION OF RIGHT TRIANGLES 
The formulae for the solution of right triangles are from 

B = 90° - A, Fig. 9, 



A 


= 90° - B, 


sin A 


a D 

— - = cos B 
h 


cos A 


b . R 

= — = sin B 
h 


tan A 


= ^=GOtB 


h 2 


= a 2 + 6 2 . 



(12) 



Case 1st 

Given A = 426.7, and A = 34° 15', to solve the triangle. 

Select from (12) the formulae which 
contain both of the given parts and 
but one of the required parts, so that 
the solution can be made without 
using a computed one, thereby di- 
minishing the unavoidable errors due 
to logarithms. In this example, 

b 




Fig. 10 



sin A = ^ , cos A = - , B = 90° - A = 55° 45', 
h h 



whence a = h X sin A, b = h X cos A. 
By Natural Functions 
a = 426.7 X .56280 = 240.15, b = 426.7 X .82659 
By Logarithmic Functions 
log h = 2.63012 log h = 2.63012 

log sin A = 9.75036 log cos A = 9.91729 



352.71. 



log b = 2.54741 



log a = 2.38048 
.-. a = 240.15* .'. b = 352.70* 

* Six figures in the result may be claimed to be correct in the first part of 
a five-place table of logarithms, but in the latter part only five figures. 

20 



SOLUTION OF RIGHT TRIANGLES 21 

Case 2d 

Given a = 5472.5, A = 32° 15' 24"; solve the triangle. 
Then B = 57° 44' 36". 

From (12), sin A =-, tan A = -, 
h o 



sin A tan A 

By Natural Functions 

10253.69, b = 54725 
.53371 .63111 



h = 54725 = 10253.69, & = 54725 = 8671.12. 



ify Logarithmic Functions 

log a = 3.73819 log a = 3.73819 

co-log sin A = 0.27269 - 10 * co-log tan A = 0.19989 



log h = 4.01088 log b = 3.93808 

.-. h = 10253.7 .-. b =8671.2 

Case 3d 

Given a = 536, h — 941; solve the triangle. 

From (12), sin A = £, & = \//? 2 - a 2 = V(A + a) (A - a). 

By Natural Functions 

sin A = ||f = .56961, b = \/l477 X 405, 

.-. ^ = 34° 43' 22", B = 55° 16' 38", 6 = 773.42. 

By Logarithmic Functions 

log a = 2.72916 log (a + h) = 3.16938 

co-log A = 7.026-*! - 10 * log (h - a) = 2.60746 

log sin A = 9.75557 log b 2 = 5.77684 

log b = 2.88842 
.-. A = 34° 43' 20", B = 55° 16' 40", b = 773.43. 

* This 10 need not be regarded if only the unit's figure in the sum of the 
characteristics is put down. Really the unit's figure is all that is needed in 
the usual logarithmic computations. 



22 PLANE TRIGONOMETRY 

Case 4th 

Given a = 643.6, b = 59.87; solve the triangle. 

From (12), tan A =-, h = Va 2 + b 2 , sin A =^.\h=-^- 
b h smA 

By Natural Functions 

tan A = ^? = 10.74996, h 2 = 417805.3769. 

59.87 

A =84° 41' 7", 5=5° 18' 53 ," h = 646.38. 

By Logarithmic Functions 

log a = 2.80862 log a = 2.80862 

co-log 6 - 8.22279 co-log sin A = 0.00187 



log tan A = 1.03141 log h = 2.81049 

A = 84° 41' 8", 5=5° 18' 52", & = 646.38 

In general a computed part should not be used to find a required 
one, but there is one exception. When either a or b is nearly 
equal to h it will be more exact to find first the remaining side 
and use that to determine its opposite angle. 

Example. Given a = 643.6, and h = 646.38; solve the triangle. 

b = V(h + a )(h- a) = 59.8844, sin B = & 

h' 
log b - 1.7773] 
co-log h = 7.18951 



log sin B = 8.96682. Hence 5=5° 18 r 57". (Compare with 

last Ex.) 

Note. When an angle is very small it may be determined by either 
its sine or tangent, since the arc, sine, and tangent are approximately 
the same; that is, 

sin x = x = tan x nearly. 

These are the same in a six-place table of natural functions to 30', and 
in a five-place table to 1°. The expression of results obtained from log- 
arithmic tables of five or six places in tenths and hundredths of a second 
is a seeming accuracy that cannot be claimed, therefore results will be 
given to the nearest second, rejecting the tenths and hundredths. One 
tenth of a minute even is greater accuracy than can be attained in meas- 



SOLUTION OF RIGHT TRIANGLES 23 

urements with the best civil engineer's transit. In geodetic and astro- 
nomic work, angles are measured to tenths and hundredths of a second, 
and a like accuracy in computations must be attained, but cannot without 
at least a seven-place table. A five-place table is sufficient for the use 
of most students. 

Area of Right Triangles 

23. From (12), a = b . cot B = b . tan A, and b = a . cot A 
= a . tan B, a = h . sin A = h . cos B, and b = h . cos A = h . sin B. 

Represent the area of the triangle by k. Then when the sides 
are given, 

k = \ ab. 

When a side and an angle are given, substitute h = a . tan B or 
a = b . tan A in the above equation, giving 
k = i b 2 tan A =U 2 tan B. 
When the hypotenuse and an angle are given, substitute 
a = h . sin A —h. cos B, or b = h . cos A = h . sin B, 
in the same equation, giving 

k = hh 2 sin A cos A = h h 2 sin B cos B, or (42) 
k = J h 2 sin 2 A =\h 2 s'in 2 B. 
Hence the three formulas for the area are, 
k = J ab. 

k = i b 2 tan A = J a 2 tan 5, (13) 

& = 1 & 2 sin 2 A = I h 2 sin 2 5. 
An Isosceles Triangle, divided by the perpendicular from the 
vertex to the base into two equal right triangles, may be solved 
when any two different parts are given, one being a side. 

EXAMPLES 

1. Given A 37° 1(7, h 8762; solve the triangle. 

Ans. a = 5293.4, b = 6982.17. 

2. Given A 34° 18', a 237.6; solve the triangle. 

Ans. b = 348.32, h = 421.64. 

3. Given a = 147, and h = 184; solve the triangle. 

Ans. A = 53° V 36", b = 110.7. 

4. Given a = 300.43, and b = 500; solve the triangle. 

Ans. A = 31°, h = 583.32. 



24 PLANE TRIGONOMETRY 

5. Given h = 365.1, and a = 348.3; solve the triangle. 

Ans. A = 72° 33', b = 109.48. 

6. Given h = 205, B = 49' 33' 6"; solve the triangle. 

Ans. a = 133.0, b = 156.0. 

7. Given h = 369.27, and 6 = 235.64; solve the triangle. 

Ans. A = 50° 20' 52", a = 284.31. 

8. Given a = 4.8199, and b = 2.6492; solve the triangle. 

Ans. A = 61° 12' 20", h = 5.5000. 

9. Given a = 340, and A = 60° 55' 51"; solve the triangle. 

Ans. h = 389.0, b = 189.0. 

10. Given B = 82° 6' 18", a = 89.32; solve the triangle. 

Ans. b = 644.11, h = 650.27. 

11. At a distance of 105 feet from the base of a tower, the angle of 
elevation of its top is observed to be 38° 25'. Find the height of the 
tower. Ans. 83.27 feet. 

12. What is the angle of elevation of the sun when a tower whose 
height is 103.74 feet casts a shadow 167.38 feet in length? 

Ans. 31° 47' 24". 

13. From the top of a house, 146 feet above the sea, the angle of 
depression of a buoy is observed to be 21° 46'. Find the horizontal 
distance of the buoy. Ans. 365.64 feet. 

14. A river 200 feet wide runs at the foot of a tower which subtends 
an angle at the edge of the opposite bank of 25° 10'. Find the height 
of the tower. • Ans. 93.97 feet. 

15. A tower stands by a river. An observer on the opposite bank 
finds the elevation to its top to be 60°. He then recedes 40 yards in a 
direct line from that point and the tower, and then finds the elevation 
to be 50°. Find the breadth of the river. Ans. 88.23 yards. 

16. The depth of a dam is 8 feet, the width of the top is 10 feet ; the 
inclination to the horizon of each side is 38° 53'. Find the width of the 
dam at the bottom. Ans. 29.85 feet. 

17. In order to find the breadth of a river, a distance AB = 96 feet 
was measured along the bank, and perpendicular to a line from A to a 
tree C on the other side. The angle ABC was measured and found to 
be 21° 14'. Find the breadth of the river. Ans. 37.3 feet. 

18. A ladder 40 feet long may be so placed that it will reach a window 
33 feet high on one side of the street, and by turning it over without 
moving its foot, it will reach a window 21 feet high on the other side. 
What is the width of the street? Ans. 56.65 feet. 



SOLUTION OF RIGHT TRIANGLES 25 

19. From the top of a hill the angles of depression to two successive 
milestones, on a straight, level road to the hill, are observed to be 5° and 
15°. Find the height of the hill. Ans. 228.63 yards. 

20. How far must a person be elevated above the surface of the earth 
in order to see an object on the surface at a distance of 15 miles, the 
radius of the earth being taken at 4000 miles? Ans. 148.5 feet. 

21. The non-parallel sides of a trapezoid are respectively 3.51 and 
7.04 meters, and if produced they will intersect at right angles. Deter- 
mine the acute angles of the trapezoid. Ans. 26 30 and 63 30. 

22. ABD is a right triangle. If AC = 329.86 feet, the angle BCD 
= 59° 12', and the angle BAD = 41° 21'. Find BD and AD. 

Ans. BD = 610.7 feet, AD = 693.9 feet. 

23. A barn is 40 by 80 feet; the pitch of the roof is 45°. Find the 
length of the rafters and the area of both sides of the roof. 

Ans. 28.284 feet, 4525.44 square feet. 

24. Given h = 8.462 feet and B = 86° 4' to find the area k. 

Ans. k = 2.456 square feet. 

25. The area of a right triangle is 4296, and A = 36° 12'. Find 
a and b. Ans. a = 79.30, b = 108.35. 

26. The area of a right triangle is 368.9 square inches, and h = 126.12 
inches. Find A. Ans. 2° 39' 41". 

27. The base of an isosceles triangle is 147 feet, and its area 2572.5 
square feet. Find its angles and sides. 

Ans. Eql. angles 25° 27' 48", alt. = 35 feet; side = 81.41 feet. 

28. Find the radius of a circle if a chord whose length is 5 feet sub- 
tends an angle of 133° at its center. Ans. 2.721 feet. 

29. The side of a regular pentagon is 20 feet. Find the length of its 
diagonal. Ans. 32.36 feet. 

30. From a square whose side is 10 feet, the corners are cut away so 
that a regular octagon is left. Find the area of the octagon. 

Ans. 82.84 square feet. 

31. Solve the isosceles triangle whose base is 1146.48 feet and vertical 
angle is 80° 36' 24". Ans. Side is 886.24 feet. 

32. If the height of an object above a level plane must be four sevenths 
(effects of curvature and refraction) of the square of its distance in miles 
to be just visible, how high must a mountain be to be just visible at a 
distance of 40 miles? Ans. 914.3 feet. 



CHAPTER III 
GENERAL FORMULA 

24. Draw the axes XX' and YY' at right angles. Draw BB" 
and B'B f " , making any acute angle x above and below XX', 
also CC" and C'C" making any acute angle y above and below 
BB" and B'B'" , x and y being taken so that their sum is less than 
90. From any point C draw CB perpendicular to OB, and BA 




Fig. 10B. 

and CD perpendicular to XX'. In a similar way construct the 
equal figures OA'B'C, OA'B"C", and OAB"'C"', one in each of 
the four quadrants. The angles B'C'E', B"C"E" , and B f "C"'E f " 
are each equal to BCE or x, because their sides are respectively 
perpendicular to those of the angle x. 

26 



GENERAL FORMULAE 27 

In the first quadrant the sin CO A equals 

. ( , N DC AB + EC AB . EC 



By (12), AB = OB sin x, and 05 = OCcosy, combining to 

AB 
eliminate OB, AB = OC sin x cos y, or — — = sin x cos ?/. Also CE 

= CB cos x, and CZ? = OC sin y, and combining to eliminate CB, 

CE =- 

above, 



7^C 

CE = OC cos x sin ?/, or — = cos x sin i/. Hence, substituting 



sin (x + y) = sin # cos y + cos £ sin y (14) 

The figures in each of the four quadrants being equal the 
ratios, DC/OC, D'C'/OC, D'C"/0C", DC"'/0C f ", are each 
evidently equal to the sin (x -f y). Furthermore, 

D'C'/OC' = sin A'OC = sin AOC = sin [(180 - x) + .(- y)] 
= sin [(90 - tf) + (90 - 2/)]. Also O'C'/OC" 
= sin A , 0C // = - sin ^OC" = - sin [(180 + x) + y] 
= - sin [(90 + x) + (90 + ?/)]. 

Likewise, D , C ,,, /OC /,/ = sin 40C'" = - sin [(360 - x) + (- y)] 
= - sin [(180 - x) + (180 - ?/)] 
= - sin[(- x) + (- ?/)]. 

Each of these by (5) to (9) can be reduced to (14). 

Hence (14) is true for the algebraic sum of any two angles of 
whatever magnitude or algebraic sign, and therefore in the formula 
either angle may be replaced by any angle whatever. 

As exercises let the student verify the above sets of values, 
thus : 
sin [(180° - x) + (- y)] = sin (180° - x) cos (- y) 

+ cos (180° - x) sin (- y), 
but by (5) and (9), sin [180 — (x + y)] = sin x cos y + cos x sin y, 
or same as (14). So for the others. 

From the above y may be replaced by — y in (14), giving after 
reducing by (9), 

sin (x — y) = sin x . cos y — cos x sin y. 



28 PLANE TRIGONOMETRY 

The same may be obtained by replacing y by 180° — y, in (14). 
Substituting in (14) for y, 90° + y gives 

sin (x + 90° + y) = sin x cos (90° + y) + cos x sin (90° + ?/), 
or by (6), cos (x +>y) = cos x cos ?/ — sin x sin ?/. 

Again, substituting for y (90° — y) in (14) gives 

sin (x + 90° — y) = sin £ cos (90° — ?/) + cos £ sin (90° — 2/), 
or by (6) , cos (x — y) = sin x sin y + cos x cos y. 

These grouped for convenient reference are, 

sin (x + y) = sin x cos y + cos x sin y (14) 

sin (x — y) = sin x cos y — cos x sin ?/ (15) 

cos (x + 2/) = cos x cos ?/ — sin x sin ?/ (16) 

cos (x — y) = cos x cos 2/ + sin x sin y (17) 

These are the four fundamental formulae of Angular Analysis, 
and they with numbers (1) to (10) inclusive of § 21 constitute the 
data from which every other required formula may be determined, 
and it is hardly possible to exaggerate the importance of learning 
how to use them to get what is needed rather than burden the 
memory too much. 

As exercises deduce (5), (6), and (7) by determining first the 
sine and cosine and then the other functions from (3) and (4); 
thus, in (15) and (17) making y equal to x gives 

sin 0° = sin x cos x — cos x sin x = 0, 

cos 0° = cos x cos x + sin x sin x = cos 2 x + sin 2 x ■ — 1. 

Then 90° and 0° being complements, 

sin 90° = cos 0° = 1, tan 90° = cot 0° = oo , esc 90° = sec 0° =1, 
cos 90° = sin 0° = 0, cot 90° = tan 0° = 0, sec 90° = esc 0° = oo . 

In (14) and (16) making x = y = 90° gives 

sin 180° = 1X0 + 0X1 = 0, tan 180° = 0, esc 180 = oo , 

cos 180° =0 X - 1 X 1 = - 1, cot 180° =- oo , sec 180 = -1. 

In (15) and (17) making successively x = 0°, 90°, 180°, 270° and 
360°, the functions of - y, 90° - y, 180° - y, 270° - y and 
360° — y result. In (i4) and (16) making successively x = 90°, 
180°, and 270° the functions of 90° + ?/, 180° + y, and 270° + y 
result. The student should verifv all these. 



GENERAL FORMULAE 29 

Functions of Two Angles. 

25. In (14), (15), (16), and (17) substitute for x and y, a and b 
throughout, and then take the sum and difference of the first and 
second results, and also the sum and difference of the third and 
fourth, giving 

sin (a + b) + sin (a — b) = 2 sin a cos b, 



sin (a + b) — sin {a — b) = 2 cos a sin 
cos (a + b) + cos (a — b) = 2 cos a cos 6, 



(18) 
cos (a — b) — cos (a + b) =2 sin a sin 
Now let a + b = x and a — b = y, whence a = % (x + y) and 
5 = i (x — 2/). Substituting these values in (18) gives 



sin x + sin y - 


■■ 2 sin J (x -f 2/) cos § (x - 


-y), • 


• (19) 


sin x — sin 2/ = 


: 2 cos i (# + 2/) sin J (x - 


-y), • 


. (20) 


cos a; + cos y = 


■■ 2 cos i (x + 2/) cos § (z - 


-2/), • 


. (21) 


cos 3/ — cos x = 


: 2 sin \ (x + y) sin J (x - 


-2/). • 


• (22) 


Divide (19) by (20), 


sin x + sin j/ _ tan J (x 
sin £ — sin y tan £ (x 


+ ?/) 


• (23) 


Divide (19) by (21), 


sin x + sin y , , , 

** = tan ^ (a: 

cos x + cos 2/ 


+ ?/)•• 


• (24) 


Divide (19) by (22), 


sin x + sin w , , / 

! * = cot J (x 

cos 2/ — cos x 


-2/). • 


. (25) 


Divide (20) by (21), 


sin x — sin y , , 

* = tan \ (x 

cos x + cos 2/ 


-2/). • 


• (26) 


Divide (20) by (22), 


sin x — sin v , , , 
* = cot \ {x 


+ y). . 


• (27) 



cos 2/ — cos x 



Divide (22) by (21), ^-^ ^- x =tan £(z + 2/) tan £(z-v).(28) 

cos 2/ + cos x 

Divide (14) by (16), and divide numerator and denominator of 
the second member by cos x cos y, giving 

, f , \ tan x + tan w /rirkX 

tan (x + y) = + ' (29) 

1 — tan x tan y 

In (29) change y to - y, tan (x - y) = tan X ~ tan v . . (30) 

1 + tan x tan y 

The reciprocals of (29) and (30) give 

, , N 1 — tan x tan y cot x cot y — 1 /on 

cot (x + 2/) = ~ = • • (31) 

tan x + tan 2/ cot x + cot 2/ 



30 PLANE TRIGONOMETRY 

cot (x - y) = 1 + tanxtani/ = cot x cot g + 1 # > (32) 
tan x — tan 7/ cot y — cot a; 
Make y = 45° in (29) and (30), and remember tan 45° = 1 

Divide (14) by (15), and divide numerator and denominator by 
cos x cos ?/. 

sin (x + y) __ tan x + tan ?/ _ cot x + cot ?/ ,„,. 

sin (a; — ?/) tan x — tan ?/ cot y — cot x ' 

Divide (16) by (17), and as above divide by cos x cos y. 

cos (x + y) = 1 — tan x tan y _ cot x cot ?/ — 1 /^v 

cos {x — y) 1+ tan x tan ?/ cot x cot y + 1 ' 

Multiply (14) by (15), also (16) by (17), and reduce by (8). 
sin (x + y) sin (x — y) = sin 2 x — sin 2 ?/ = cos 2 y — cos 2 x. (36) 
cos (x + 2/) cos (x — y) = cos 2 x — sin 2 ?/ = cos 2 ?/ — sin 2 x. (37) 
Make a = 30° and b = ?/ in (18) give 

sin (30° + y) + sin (30° - y) = cos_y. See (10) . . (38) 
sin (30° + y) - sin (30° - y) = V3 sin ?/.... (39) 
cos (30° + y) + cos (30° - y) = V3 cos ?/.... (40) 
cos (30° + y) - cos (30° - y) = sin y (41) 



Functions of Twice an Angle. 

26. Making y = x in (14) and (15), and dividing each by 
= sin 2 x + cos 2 x, 

2 sin x cos x 2 tan x 



sin 2 x = 2 sin x cos x = 

sin' x + cos 4 x 1 + tan 4 x 

O rt^.+ /v. 

(42) 



cot 2 X + 1 

cos 2 x — sin 2 x 1 — tan 2 x 



cos 2 x = cos 2 x — sin 2 x = 

cos 4 x -f- sin 4 x 1 + tan 4 x 

(43) 



cot 2 X + 1 
Substitute cos 2 x = 1 — sin 2 x, and then sin 2 x = 1 — cos 2 x in 

(43) and reduce, 

2 sin 2 x = 1 — cos 2 x. .*. sin x = V^ (1 — cos 2 x) (44) 

2 cos 2 x = 1 + cos 2 x. .*. cos x = \/\ (1 + cos 2 x) (45) 



GENERAL FORMULAE 31 

si n 2 t 
Reverse (42) and divide by (45), tan x = — — . (46) 

1 + COS L X 

Divide (44) by (42) reversed, tan x - 1 ~. C ° S 2 X . (47) 

sm £ x 



Vl — cos 2 z /.qv 
Divide (44) by (45), tan x = • (48) 

VI + cos 2 a: 

Make y = x in (29) and (30), tan 2 x = i 2 ^ n X 2 ■ (49) 

1 — tan- x 

and cot 2 x = CO f g ~ * . (50) 

2 cot x 

Adding (42) to and subtracting it from 1 = cos 2 x + sin 2 x and 

reducing, 

1 + sin 2 x = (cos x + sin x) 2 (51) 

1 — sin 2 x = (cos x — sin x) 2 (52) 

Extract square root of (51) and (52), add and subtract results. 
2 cos x = ± Vl + sin 2 x ± V 1 - sin 2 x . . (53) 
2 sin x = ± Vl + sin 2 x + Vl - sin 2 x . (54) 

Divide (43) first by (51) and then by (52) and reduce, 
cos 2 x cos x — sin x 1 — tan x 



1 + sin 2 x cos x + sin x 1 + tan x 
cos 2 x _ cos x + sin x _ 1 + tan x 
1 — sin 2 x cos x — sin x 1 — tan x 
Divide (42) and (43) each by 1 = cos 2 x + sin 2 x. 



(55) 
(56) 



. 2 sin x cos x 2 tan x /e ~ 

sm 2 x = — n ; ,:,_»„ = , , ,_„, „ • • • (57) 



cos 2 x + sm 2 x 1 + tan 2 x 
cos 2 x sin 2 x _ 1 — tan 2 x 
cos 2 x + sin 2 x 1 + tan 2 x 



cos 2x = 52^ gigj_s = 1-tan'a^ _ _ (5g) 



Functions of Half an Angle. 

27. Substitute x for 2 x and ^ x for x in § 26, the following will 

result : 

From (42), sin x = 2 sin ^x cos ^x = 2 tan i- x = 2 cot i x _ 

1 + tan 2 J x 1 + cot 2 J x 

(59) 

From (43), cos x = cos 2 J x - sin 2 J x = - ~ tan ' 2 g 

1 + tan 2 y x 

oof 2 1 r — 1 
= CQI 2 ,T - 1 • (60) 

COt 2 $3 + 1 \ 



32 PLANE TRIGONOMETRY 

From (44), 2 sin 2 J x = 1 — cos x. .'. sin \ x = Vj (1 — cos z). 

(61) 

From (45), 2 cos 2 J x = 1 + cos x. .*. cos J x = V£ (1 + cos z). 

. (62) 

AlsoV2cos^ = Vl + cosx^J 1 -™* 2 ^ , SinX • 

V 1 — COS £ VI — COS X 

(63) 
From (46), (47), (48), 

+„~ i «. sin -a; 1 — cos x _ , /l — cos a? /ftil * 

tan |x= — = : 1/ . (64) 

1 + cos x sinx V 1 + cos x 

From (49), (50), 

tan x = 2tan t* ,cots= cot2 ** = 1 . . . (65) 
1 — tan 2 ^ x 2 cot ^ x 

From (51), 1 + sin x = (cos J x + sin J x) 2 . .... (66) 

From (52), 1 — sin x = (cos \ x — sin \ x) 2 (67) 

From (53), 2 cos i a; = ± Vl + sin a; ± Vl - sirTx. . (68) 

From (54), 2 sin J a; =" ± Vl + sin a; q= Vl - sin x. (69) 

From (55), (56), 

cos x _ 1 — tan \ x , cos a; __ 1 + tan \ x ] , 7 ~>. 

1 + sin as 1 + tan J x ' ' 1 — sin x 1 — tan ^ a; 

From (57), (58), 

2 tan \ x -, 1 — tan 2 \ x /- 1N 

sin a; = \ , and cos x = -f — . (71) 

1 + tan 2 i a; 1 + tan 2 J a; 

Functions of Three Angles, Three Times and Four Times an Angle. 

28. sin (x + y + z) = sin (x + ?/) cos z + cos (a: + y) sin 2, or 
sin (x + y + z) = sin x cos y cos z + cos a: sin y cos 2 

+ cos x cos 7/ sin z — sin a? sin y sin 2. . . . (72) 
By changing z to — z in (72) and reducing by (9) , 

sin (x + y — z) = sin x cos ?/ cos z + cos a: sin y cos 2 

— cos x cos ?/ sin z + sin a; sin y sin z. . . . (73) 
By changing i/ to — 3/ in (73) and reducing, 

sin (x — y — z) = sin x cos y cos z — cos a: sin y cos 2 

— cos x cos ?/ sin 2 — sin a: sin y sin 2. . . (74) 
cos (x + 2/ + z ) = cos (a: + ?/) cos 2 — sin (x + ?/) sin 2, or 
cos (x + y + 2) = cos x cos t/ cos 2 — sin x sin 1/ cos 2 

— sin a; cos y sin 2 — cos x sin ?/ sin 2. . . (75) 



GENERAL FORMULAE 33 

By changing z to — z in (75) and reducing, 
cos (x + y — z) — cos x cos y cos z — sin x sin y cos z 

+ sin x cos ?/ sin z + cos x sin ?/ sin z (76) 

By changing y to — 2/ in (76) and reducing, 

cos (x — y — z) = cos x cos ?/ cos z + sin x sin i/ cos z 

+ sin x cos y sin z — cos x sin ?/ sin z (77) 

Divide (72) by (75), and then divide numerator and denomi- 
nator of the second member by cos x cos y cos z. 
, , v _ tan x + tan ?/ + tan z — tan x tan ?/ tan z /-~. 

1 — tan x tan ?/ — tan x tan z — tan y tan z 
Making x = y = z'm (72), (75), and (78), 

sin 3 x = 3 cos 2 x sin a: — sin 3 x = sin x (3 — 4 sin 2 x) 

= 3 sin x — 4 sin 3 x (79) 

cos 3 x = cos 3 x — 3 sin 2 x cos £ = cos 3 x — 3 cos x (1 — cos 2 x) 

= 4 cos 3 x — 3 cos x (80) 

. r» o ian x ian x /o-«\ 

tan3 * = l-3tan 2 x (81) 

By (42), (43), 

sin 4 x = 2 sin 2 x cos 2 x = 4 sin x cos x (2 cos 2 x — 1) 

= 8 sin x cos 3 x — 4 sin x cos x (82) 

cos 4 x — cos 2 2 x — sin 2 2 x = (2 cos 2 x — I) 2 

— 4 (1 — cos 2 x) cos 2 x = 8 cos 4 x — 8 cos 2 x + 1. (83) 



EXERCISES. 

1. Prove sin 75° = * + ^- 3 = .9659, sin 75° = sin (45° + 30°). 

See (14) 

2. Prove cos 75° = 3 T— = -2588. See (16) 

\/3 - 1 

3. Prove sin 15° = ■ -?=- . See (15) 

1 _l_ V3" 

4. Prove cos 15 = 7=- See (17) 

5. Prove tan 75° = 2 + V3. 

6. Prove tan 15° = 2 - VS. 

7. If sin x = .8 and sin ?/ = .6, find the value of sin (x + t/) and 
cos (x + y). 



34 PLANE TRIGONOMETRY 

8. Prove = sin <* + y> ± sin {x ' y) = tan*. 

cos {x + y) + cos (x - y) 

A -r, tan a; cot y + 1 , , , N 

9. Prove 2 = tan (x + w). 

cot y — tan a; 

10. cos (x - 45°) - sin (x + 45°). 

11. If tan x = \ and tan y = \, prove tan (x + y) = f , and tan (a; — i/) 

_ 2 
~~ 9- 

12. If tan x = f and tan ?/ = rl, prove tan (# + y) = 1. 

13. If tan 2 a: = 2, find the functions of x. See (42) 

14. Find the functions of 25° in terms of 50°. See (44), (45), (64) 

15. If x + y + z = 90, prove tan 2 = 1 ~ tan x tan y . See (5), (31) 

tan x + tan y 

16. sin 3 x + sin 5 x = 2 sin 4 a: cos x. See (19) 

17. Prove sin 7 x — sin 5 x =2 cos 6 a: sin x. See (20) 

18. Prove cos 5 x + cos 9 x = 2 cos 7 a: cos 2 x. 

19. Prove cos 4 x - cos 5 z = 2 sin | x sin ^ a\ 

20. Prove cos (60 + x) + cos (60 - 3) = cosje. See (10) 

21. Prove cos (45 + x) + cos (45 - x) = V2 cos x. See (10), (21) 

22. Prove sin 2 z cos 2 z = J (1 - cos 4 jc). See (42), (44) 

23. Simplify sln f * " sin \ - . See (20), (21) 

cos 5 x + cos 3 £ 

24. Simplify sin * ± sin \ X . See (19), (23) 

cos x — cos 5 x 

oc -r, sin 5 x + sin £ , 

25. Prove = tan 3 x. 

cos 5 x + cos x 

nn -r, cos £ + cos 3 x cos 2 # 

26. Prove — — — - = — . 

cos 3 x + cos 5 x cos 4 a; 

27. Prove sec 2 x esc 2 z = sec 2 x + esc 2 a\ 

1 + sin x — cos x , 1 „ 

28. Prove : : = tan \ x. 

1 4- sin x + cos x 

nn _, cos 3 x + sin 3 # 2 - sin 2 a; 

29. Prove — ; = ~ • 

cos x + sin x 2 

30. Prove cos 4 x — sin* x = cos 2 x. 

31. Prove sin 3 x + cos a: = 2 sin (45 + x) cos (45 



32 





See 


(64) 




See (7), 


(42) 


2 


#). 

See (5), 

cos a; 


(19) 




1 - sin x 

See (48) 


,(7) 



-r, j. tATt , i n 4 /l + sinz 1 + s 

Prove tan (45 + \ x) = 1/ = — = — 

V 1 - sin x cos 

«« -r. x . \ sm fa ± V) 

33. Prove tan a; ± tan y = '— . 

. cos a; cos y 

34. Prove sin 3 a: + 2 sin 5 a; 4- sin 7 x = 4 sin 5 a: cos 2 x. See (19) 

35. Prove cos 3 a: + 2 cos 5 a; + cos 7 a: = 4 cos 5 x cos 2 x. See (21) 



GENERAL FORMULAE 35 

-. -p, tan (x + y) - tan x , 

36. Prove — — — — :- = tan y. 

1 + tan {x + y) tan x 

«rr t> /tana: + 1\ 2 1 + sin 2 x ,__. ,_ fiN 

37. Prove ( = ^—7— . See (5/) -*- (06) 

\tana; - 1/ 1 - sin 2 a; 

38. sin 50° + sin 10° = sin 70°. See (19), (5) 

39. Prove sm . 3 x ~ cos3x = 2 sin 2 x - 1. See (79), (80) 

sin x + cos x 

40. Prove sin * + sm 2 * = tan 3. See ( 46 ) (3) 

1 + cos x + cos 2 x 

„, r> * 100 sin 33° + sin 3° 

41. Prove tan 18° 



cos 33° + cos 3° 

42. If x + y + z = 180°, prove tan x + tan y + tan z = tan a; tan // 
tan 2. See (78) 

43. li x + y + z = 180°, prove sin x + sin 7/ 4- sin 2=4 cos J z cos \ 
ycos^z. See (59), (19) 

44. Find the values of x when sin 3 £ + sin 2 a; + sin x = 0. 

See (79), (42) 

45. Prove cos x + cos 3 x + cos 5 x + cos 7 a; = 4 cos a: cos 2 z cos 4 a;. 

See (21) 

46. Prove 2 sin 2 x sin 2 y + 2 cos 2 a: cos 2 1/ = 1 + cos 2 a; cos 2 y. 

See (44), (45) 

47. Prove cos 2 * ~ cos 4 a; = tan 3 x% g ee (22 ), (20) 

sin 4 a: — sin 2 x 

48. Prove 4 sin a: sin (60 - x) sin (60 + x) = sin 3 a:. See (79) 

49. Prove tan 2 } a; = 2sm * ~ s ' n2x . See (42), (64) 

sm 2 a: + 2 sin x 



/l + c\l 
50. If tan \ x = ( I tan £ y, prove cos a; 



cos y — c 



1 — c . cos i/ 

See (48) 

ki T f 1 oao sin a- + sin v - sin z , . , 

51. Hx + y+ z = 180 , prove — : — - — — - — = tan * x tan * y. 

sin x + sm y + sin z 

See (19), (42) 

52. If x + y + z = 180°, prove cot | £ + cot $■ ?/ + cot ^ z = cot £ a; 
cot £ ?/ cot £ z. See (31) 

53. If x + y + 2 = 180°, prove sin 2 x + sin 2 y + sin 2 z = 2 + 2 cos x 
cos y cos z. See (75) 

54. If x + y + z = 180°, prove tan ^ x tan ^ ?/ + tan £ ?/ tan £ + 
tan £ 2 tan \ x = 1. See (78) 

55. If z + y + z = 180°, prove cos 2 a; + cos 2 3/ + cos 2 z + 4 cos x 
cos ?/ cos z + 1 =0. [Ex. 53.] 

56. Prove tan (45° + y) - tan (45° - y) = 2 tan 2 y. See (29), (30) 



36 PLANE TRIGONOMETRY 



57. Prove sin (45° -**)•+ cos (45° -\x) 



VI - cos a; 

See (5), (21), (63) 

58. Prove sin a; sin (?/ — z) + sin ?/ sin (z — x) + sin z sin (x — y) = 0. 

59. Prove cos z sin (y — z) + cos ?/ sin (z — x) + cos z sin (x — y) = 0. 

60. Prove the values x in sin 3 z = 4 sin x sin 2 z sin 4 x, are 0°, 20°, 
40°, 80°. See (79), (43) 

e-i -d tan 3 x , cot 3 x 2 — sin 2 2 # „ , Ans , ox 

61. Prove — — + ^^^ = ^ • See (42), (8) 



Inverse Trigonometric Functions. 

29. The equations sin x = a, cos y = b, tan 2 = c, signify that 
x is an arc whose sine is a, y an arc whose cosine is b, z an arc 
whose tangent is c, and they may be written in the inverse forms, 

x = sin -1 a, y = cos -1 b, z = tan -1 c, also u = log -1 d. 

These are inverse trigonometric junctions, and are read, x equals 
an arc whose sine is a, y equals an arc whose cosine is b, z equals 
an arc whose tangent is c, and u equals a number whose loga- 
rithm is d. The customary method abbreviates, thus, x = arc 
sin a, y = arc cos b, z = arc tan c, and u = anti-logarithm d, or 
written as above. The minus one ( — 1) is in no sense an expo- 
nent, only a symbol. 

The following are evident : sin x = J, x = sin -1 J, sin x = sin 
(sin -1 J) = J, and sin -1 (sin x) = x = sin -1 J. The last reads : 
arc sin (sin x) equal to x equal to an arc sin J. 

To find one value of the arc whose tangent is tan -1 J + tan -1 J. 

Let x = tan -1 \ and y = tan -1 J. Then tan x = \, tan 2/ = J, 

and tan (x + y) = * + 7 - = 1 = tan 45°. 

1-1X1 

Hence 45° = tan" 1 J + tan" 1 J. 

Prove the following for the least positive values. 

1. sin -1 § = cos -1 3 = tan -1 f . Assume sin x = f . 

2. sin- 1 \ = cos-^Vf =cot- x V3. 

/ ■ a 

3. sin -1 a = cos -1 VI — a 2 = tan /- — — % • 

Vl-a 2 

a , 1 , ± 1 *. x m, + rri2 

4. tan -1 m x 4- tan -1 ra 2 = tan -1 — • 

1 — m, wis 



GENERAL FORMULAE 37 

5. 2 cos- 1 a = cos- 1 (2 a 2 ~ 1). See (43) 

6. sin- 1 1 + sin- 1 T V + sin" 1 |f = sin~ l 1. See (72) 

7. sin- 1 m + cos -1 m = 90°, or - . 

8. Find the value of tan (cos- 1 X- sin" 1 /-)• Ans. a ~ 

9. Find the value of tan (2 tan" 1 2). 

10. Find the value of cos (3 sin- 1 | ^5). 

11. Prove 4 tan" 1 £ - tan- 1 ^ = 45°, or j • 

12. Prove 2 sin- 1 1 - sin" 1 U = 2 cos" 1 1| • 



VI — a 2 

13. Show that tan -1 + sin -1 a is a constant for all values of a. 

a 

14. Show tan -1 x = tan -1 \ + tan -1 j\ gives x = \. 

15. If x = cos -1 a - sin -1 6, find x = a Vl - 6 2 + 6 Vi - a 2 

16. If x = 2 tan- 1 £ + tan- 1 \, find a; - 45°. 



CHAPTER IV 



SOLUTION OF OBLIQUE TRIANGLES 

Formulas for the Right Triangle Method 

30. All cases of oblique plane triangles can be solved by the 
formulae for right triangles and two additional ones deduced 
from them. 

Case 1st. Given a Side and Two Angles 
Two angles of a triangle given all the angles are known. 
Suppose AC and the angles known in Fig. 11. Then by (12), 
BD = c. sin A, and BD = a. sin C. Hence 

sin C 



c. sin A = a. sin C. or c = a 



a = 



b = a 



sin A 
sin B 
sin B 
sin A 
sin C 
sin A 



= c 



= b 



sin 


A 


sin 


A 


sin 


C 


sin 


B 


sin 


C 


sin 


C 



sin A 
sin C 



sin B 



(84) 



Case 2d. 




Given Two Sides and an Angle Opposite One of Them. 
Given the angle A and the sides 
a and c. By (12), AD = c . cos 
A in the right triangle ADB. In 
the right triangle DBC, a is known 
and DC = b — c . cos A, or c . cos 
A — b, which therefore may be 
solved. The several possible solu- 
tions in this case will be given in § 40. 

Case 3rd. Given Two Sides and Their Included Angle 

In Fig. 11, given b and c and the angle A. BD is perpendicular 
to AC. Then by (12), BD = c . sin A, and AD = c . cos A. In 



Fig. i: 



SOLUTION OF OBLIQUE TRIANGLES 39 

the right triangle DCB, the sides CD = b - c . cos A, or c . cos A 
— 6, and a being known, the solution may be completed. 

Case 4th. Given All the Sides a, b, c 
Suppose b to be the longest side. Draw the perpendicular BD 
to it from the opposite angle. Then 

BD 2 = AB 2 - AD 2 = EC 2 - DC 2 , whence 
AB 2 - BC 2 = AZP - DU 2 = (AD + DC) (AD - DC), or 
C 2 _ a 2 = b (40 _ DC), whence 

AD-DC= C —^ (85) 

b 

Substituting for DC = b - AD and solving for AD, gives 

^. y + ^-tf ^tojc. tf + y-* . 

26 26 

The triangle is now divided into two right triangles with a 
hypotenuse and a side in each known, and its solution may be 
completed. 

Formulae for the General Method 

31. From § 30 is derived 

-^— = _A_ = ^- , or 
sin A sin B sin C 

The sides of a plane triangle are proportional to the sines of their 
opposite angles. 

32. The first equation of (84) gives - = sin n , whence by 

6 sin B 

composition and division, the following 

a + 6 _ sin A + sin B 
a — b sin A — sin B 

By (23), Sin \ + Si " | = * a " |44 + gj , giving 

sin A — sin 5 tan \ (A — B) 

a + 6 tan 4 (A + £) , w . DN a — b , t , A . D \ 

^ = : rh m ' or tani (A -5)= — — -tan \ (A + B). 

a - 6 tan ^ (A — B) a + b 

(86) 

Hence, 

The sum of two sides of a triangle is to their difference as the 
tangent of the half sum of their opposite angles is to the tangent of 
their half difference. 



40 PLANE TEIGONOMETRY 

33. In Fig. 11, § 30 DC = a . cos C, CD = a . cos (180 - C) 

= — a . cos C, 
and AD = c . cos A, therefore the two values of 6 are, 

AC = 6 = AD + DC, 

AC = 6 = AT) - CD, 
and in both, b = a . cos C + c . cos A,^ 

likewise, c = 6 . cos A + a . cos B, I . . . . (87) 

likewise, a = b . cos A + c . cos Bj 

Hence, 

Any side of a triangle is equal to the sum of the projections of the 
other two sides upon it. 

Also Fig. 11, tan A = a - sm C — (88) 

b — a . cos C 

34. From the first equation of (87), 

a . cos C = b — c . cos A, 
and from (84), a . sin C = c . sin A. 

Squaring, adding, and reducing these by (8), gives 
a 2 (sin 2 C + cos 2 C) = 6 2 - 2 6c . cos A + c 2 (sin 2 A + cos 2 A), or 
a 2 = b 2 + c 2 — 2 6c . cos A,^j 
b 2 = a 2 + c 2 - 2 ac . cos 5, 1 . . . . (89) 
c 2 = a 2 + b 2 - 2 a& . cos C J 
Hence, 

T/ie square of any side of a triangle is equal to the sum of the 
squares of the other two sides diminished by twice their product into 
the cosine of their included angle. 

Note. From this proposition the whole subject of plane trigonometry 
can be developed. See Clarke's Geodesy, p. 38. As (105) of spherical 
trigonometry is determined from (89), it is the fundamental formula of 
the whole subject. 

35. From the first of equations (89) is found 

A b 2 + c 2 - a 2 

cos A = — . 

2 6c 

Subtracting both sides from unity, 

1 a i b 2 + c 2 - a 2 2 6c - 6 2 - c 2 + a 2 

1 — cos A = 1 — = — 

2 6c 2 6c 

_ a 2 - (6 - c) 2 

2 6c 



SOLUTION OF OBLIQUE TRIANGLES 



41 



By (44), 1 — cos A = 2 sin 2 £ A. Substituting and extracting 
square root, 

sin | A = y- 



(b-cy 



4 6c 



-V 



(a 



6 + c) (a + b - c) 
4 be 



Let a + 6 + c = 2 s, whence a — 6 + c = 2 (s — 6) and a + 6 
- c = 2 (s - c). 

Substituting and reducing, 

^- 6) (s - c) ) 



sin £ A = 

sin i B = y/- 

. /(s ~ «) (s ~ c) 



_^ 

(s — a) (s — c) 
ac 



ab 



(90) 



sin i C = 

Hence, 

The sine of one half of an angle of a plane triangle is equal to the 
square root of the product of the half sum of the sides less one adja- 
cent side, by the half sum less the other adjacent side, divided by the 
product of the adjacent sides. 

36. Adding both sides of the equation for cos A to unity gives 



1 + cos A = 1 



b 2 + c 1 



(b + c) 2 



2 be 2 be 

But by (45), 1 + cos A = 2 cos 2 £ A, which substituted and 
reduced as above gives 

cos h A 



\M 



+ c) (b + c - a) 



4 6c 



Substituting as in § 35 gives 

cos £ A = i/ 6 

cosi£-i/ 



s ( s — a) 



cos * C = 



V 

v/ 



ac 



s (s — c) 



ab 



(91) 



Hence, 

The cosine of one half of an angle of a plane triangle is equal to 
the square root of the product of the half sum of its sides by the half 
sum less the opposite side divided by the product of the adjacent 
sides. 



42 PLANE TRIGONOMETRY 

37. Dividing (90) by (91) gives 



tan i A 


- \i" 


s (s — a) 




tan \ B 
tan \ C 


-\/<- 
-^ 


- a) (s - 
s (s - b) 

- a) (s - 
s (s — c) 


c) \ 

> 

b) 



(92) 



Hence, 

The tangent of one half of an angle of a plane triangle is equal 
to the square root of the product of the half sum of its sides less one 
adjacent side, and the half sum less the other adjacent side, divided 
by the half sum multiplied by the half sum less the opposite side. 
Area of Oblique Triangles 

38. Case 1. Given two sides and their included angle, find the 
area. 

By Fig. 11, of § 30, BD = c . sin A, and by the rule, the area 
k equals half the base multiplied by the altitude. 

k = \ be . sin A = \ ac . sin B = \ ab . sin C. . . (93) 

Case 2. Given one side and the angles, find the area. 

By Case 1, k = J be . sin A, and by (84), b = c ' sm „ B , hence 

sin C 

i _ ! „ 2 sm A sm B -- i ^2 sm A sm Q _ i 2 sin B sin C , QA . 

K — 2 C ; — — 2 : — — f a ; - . (\M:) 

sm C sm B sin A 

Case 3. Given all the sides, find the area. 

By Case 1, k = \ be . sin A = be . sin J A cos \ A. Then substi- 
tuting sin J A from (90) and cos \ A from (91) gives 



k = be y A S - Ye" ~ ^ X \/' 



be 



a) 
- 1 , or 



k = Vs (s - a) (s - b) (s - c) (95) 

Remark. The formulae of §§ 30 to 38 are all that are neces- 
sary in the solutions of oblique triangles. There are many 
formulae showing curious relations between the angles of tri- 
angles, a few of which are given in Exercises Nos. 42, 43, 51, 52, 
53, 54, 55, § 29. There are also several showing the relations of 
the sides and angles of triangles which are good exercises in 
angular analysis. 



SOLUTION OF OBLIQUE TRIANGLES 43 

Solution of Oblique Triangles 

39. Case 1. Given one side and two angles, as a, B, C. 

A = 180° -B- C, and from (84), b = a ' . sin A B , and c= a ' sinC - 

sin A sin A 

Example. Given a = 1764.3 feet, B = 66° 39', and C = 94° 54'. 
A = 180° - 66° 39' - 94° 54' = 18° 27'. 

By Natural Functions 
b = 1764.3 X .91810 = 
.31648 
1764.3 X .99634 = 55544 
.31648 

B?/ Logarithmic Functions 



log 6 = log a + log sin 


B + colog sin A — 10.* 


log c = log a + log sin 


C + colog sin A - 10.* 


log a = 3.24657 


log a = 3.24657 


log sin B = 9.96289 


log sin C = 9.99841 


log sin A = 0.49966 


colog sin A = 0.49966 


log b = 3.70912 " 


log c = 3.74464 


b = 5118.2 feet. 


c = 5554.4 feet. 



40. Case 2. Given two sides and an angle opposite one of them, 
b, c, B. 

From (84), sin C= -^^ — ; in which, since sin C = sin (180° 
o 

— C), C may have either or both these values. 

First. If B is acute and b is less than c and greater than 
c . sin B = AD, there will be two solutions, since ABC and ABC 
contain all the given parts, Fig. 12, I. 

If b = C-. sin B, the triangle will be right, and there can be but 
one solution, Fig. 12, II. 

If b is less than c . sin B, the triangle and solution are impos- 
sible, Fig. 1 1, III. 

If b is greater than c there can evidently be but one solution, 
Fig. 12, IV. 

* 10 is always omitted by using only the unit's figure in sum of char- 
acteristics. 



44 PLANE TRIGONOMETRY 

Second. If B is obtuse, b must be greater than c, that there 
may be one solution, Fig. 12, Y. 

Example. Given b = 379.41 ft., c = 483.74 ft., and 5 = 34° 11'. 




Fig. 12. 



sin C 



By Natural Functions 
_ 483.74 X .56184 = 716335 . ... Q « 45° 45' 10", 



379.41 
C'= 134° 14' 50", 
whence A = 100° 3' 50", A' = 11° 34' 10' 



a = 



S^O^S^ = 664 9 feet> 
.56184 

J79 1 4]_>C20p56 =1 35 > 43feet. 
.56184 



SOLUTION OF OBLIQUE TRIANGLES 45 

By Logarithmic Functions 

log sin C = log c + log sin B + colog b, 

log a - log b + log sin A' + colog sin B, 
log a = log 6 + log sin A + colog sin B. 
log c = 2.68462 log 6 = 2.57911 log 6-2.57911 

log sin B = 9.74961 log sin A' = 9.99326 log sin A' = 9.30223 

colog b = 7.42089 colog sin B = 0.25039 colog sin B = 0.25039 

log sin (7 = 9.85512 log a = 2. 82276 log a' = 2. 13173 

C =45° 45' 10" a = 664.9 ft. a' =135.43 ft. 

C'=134° 14' 50" 

41. Another solution of Case 2 by natural functions may be 
had from (89). 

b 2 = a 2 + c 2 — 2 ac . cos 5, whence 
a 2 — 2 ac . cos 5 = b 2 — c 2 , or 



a = c . cos 5 ± V6 2 — c 2 + c 2 . cos 2 B = c . cos 5 
± V& 2 - c 2 sin 2 !?. 

Evidently there is but one solution when b = c . sin B, and 
none when 6 is less than c . sin 5. 

When the two angles C and C are nearly 90° it is more accurate 
to find AD = c .sin B, and determine the angle CAD = CAD 
by the second method in Case 4 of right triangles, § 22; that is, 
find the value of CD = CD ±Vb 2 - c 2 sin 2 B 



= V (& + c . sin 5) (6 - c . sin B), 

then the sin CAD = sin CAD = C -^ , and C = 90° - CAD and 

o 

C = 90° + CAD. 

43. Case 3. Given two sides and the angle between them, b, c, A. 
B + C = 180° - A, and from (86), 

tan i (B - C) = ^-^ X tan £ (5 + C) = ^-=-^cot \ A. 
6 + c b + c 

Example. Given b = 456.12 chains, c = 296.86 chains, 
A = 74° 20'. 



46 PLANE TRIGONOMETRY 

By Natural Functions 

tan i (B - C) = 4t^H X L3190 = - 2789 8. 

752.98 

i (5 - G) = 15° 35' 17", 

J (5 + C) = 52° 50', 

whence 5 = 68° 25' 17", and C = 37° 14' 43". 

a = **.12x. 98285 = 472.27 chains. 
.92991 



By Logarithmic Functions 

log tan \ (B — C) = log (6 — c) + log cot \ A + colog (6 + c), 
log a = log b + log sin A + colog sin B. 

log (6 - c) = 2.20211 log b = 2.65908 

log cot \ A = 0.12026 log sin A = 9.98356 

colog (b + c) = 7.12322 colog sin 5 = 0.03155 

log tan J (5 - C) = 9.44559 log a = 2.67419 

J- (B - C) = 15° 35' 19", B = 68° 25' 18" a = 472.27 ch. 

C = 37° 14' 42" 

If only the third side is required, it may be obtained by (89), 
thus: 



a = Vb 2 + c 2 — 2 be . cos A 



= \/208045.45 + 88125.86 - 73128.87, or 
a = 472.27 chains. 



43. Case 4. Given all the sides, a, b, c. 

The solution may be made by (90), (91), or (92), but the pre- 
cision of (92) is the greatest; and when all the angles are required 



SOLUTION OF OBLIQUE TRIANGLES 



47 



it makes less work. It, however, may be placed in a more con- 
venient form, thus: 



tan ±A = v / (s -*)(«- c) = ^/ (s - a) (s - 6) (s - c) 



V 



V s (s — a) 

1 /(s - a) (s - 6) (s - c) 

- a V 



(s - a)' 



s — a ▼ s 

Placing the radical equal to r (see § 46) gives 

1 



tan -A = 



tani-S = 



tani-C 





r 


s 


— a 
r 


s 


- b 
r 


s 


— r 
j 



(96) 



With natural functions this case is easily solved by (89) : 



cos A = 



b 2 + c 2 



2 be 



Example. Given a = 352.25, b = 513.27, c = 482.68 yards. 

By Natural Functions 

(513.2 
cos 



A (513.27) 2 + (4S2.68) 2 - (352.2S) 2 nw „ A 

A = - — = ./olo4. 

2 X 513.27 X 482.68 

whence A = 41° 16' 56". 



By Logarithmic Functions 
log r = h {log (s — a) + log (s — 6) + log (s — c) -f colog s }, 

log tan \ A = log r + colog (s — a), 
s = 674.10. s - a = 321.85. s - 6 = 160.83. s - c = 191.42. 
log (s - a) = 2.50766 log r = 2.08365 

log (s - b) = 2.20637 colog (s - a)= 7.49234 
log (s - c) = 2.28199 log tan \ A = 9.57599 
colog s = 7.17128 | A = 20° 38' 28" 

2 log r = 4.16730 A = 41° 16' 5 6" 

log r = 2.08365 log r = 2.08365 

colog (s - b) = 7.79363 colog (s - c) = 7.71801 
log tanO = 9.87728 log tan J C = 9.80166 

i £ = 37° o' 38" * C = 32° 20' 56" 



B 



'4° or 16" 



C = 64° 4r 52' 



A + B + C = 1S0° 0' 4" (error from tables) 



48 PLANE TRIGONOMETRY 

This case is solved more easily by (85), 4th case of right tri- 
angles, § 30, than in any other way. 

Areas of Oblique Triangles 

44. Example. Given b = 14.135 chains, c = 23.667 chains, 
A = 33° 17' 13". 
From (93), k = i 6c sin A. 

log k = log b + log c + log sin A + colog 2. 
log b = 1.15030 
log c = 1.37414 
log sin A = 9.73944 
colog 2 = 9.69897 

logfc = 1.96285 .*. k = 91.802 square chains. 
Example. Given a = 18.063 chains, A = 96° 30' 15", 
B = 35° 0' 13". 

From (94), k = a 2 : — - — , whence 

2 sin A 

log k = 2 log a + log sin B + log sin C + colog sin A + colog 2. 
C = 180° - 96° 30' 15" - 35° 0' 13" = 48° 29' 32". 
log a 2 = 2.51358 
log sin B = 9.75863 
log sin C = 9.87440 
colog sin A = 0.00280 
colog 2 = 9.69897 
log k = 1.84838 .*. fc = 70.531 square yards. 
Example. Given a = 63.89 yards, b = 138.24 yards, 
c = 121.15 yards. 



From (95), k = Vs (s — a) (s — b) (s — c), whence 

log k = i Jlog s + log (s - a) + log (s - b) + log (s - c)}. 
s = 161.64, s - a = 97.75, s - b = 23.40, s - c = 40.49. 
log s = 2.20855 
log (s - a) = 1.99012 
log (s - b) = 1.36922 
log (s - c) = 1.60735 
2\ogk = 7.17524 
log k = 3.58762 .'. k = 3869.2 sq. yards. 



SOLUTION OF OBLIQUE TJBIANGLES 49 

EXAMPLES 

1. Given a = 795 feet, A = 79° 59', and B = 44° 41'; 

find b = 567.688 feet, and c = 663.986 feet. 

2. Given a = 820 meters, A = 12° 49', and B = 141° 59'; 

find 6 = 2912.20 meters, and c = 1573.89 meters. 

3. Given c = 1005 feet, A = 78° 19', and B = 54° 27'; 

find a = 1340.64 feet, and b = 1113.82 feet. 

4. Given a = 6412 feet, A = 70 55', and C = 52° 9'; 

find b = 56S5.9 feet, and c - o357.5 feet, 

5. Given a = 999 meters, B = 37° 58', and C = 65° 2'; 

find b = 630.77 meters, and c = 829.48 meters, 

6. Given A = 50°, a = 119, and 6 = 97; 

find B = 38° 38' 19", C = 91° 21' 41", c = 155.3. 

7. Given A' = 41° 13', a = 77.04 feet, 6 = 91.06 feet; 

find B = 51° 9' 6", C = 87° 37' 54", and c = 116.824 feet, 
and B' = 128° 50' 54", C = 9° 56' 6", and c'= 20.173 feet. 

8. Given C = 30° 40' 35", a = 221, and b = 149; 

find A =110° 0' 58", B = 39° 18' 27", and c = 120. 

9. Given a = 5134, b = 7268, and c = 9313; 

find A = 33° 15' 38", B = 50° 56', and C = 95° 48' 22". 

10. Given 5 = 68° 10' 24", a = 83.856, and b = 83.153; (see Art. 41) 

find |^, = Yi0° 35' 00", c= 60 ' 410 ' and c ' = 1>944 - 

11. Given a = 565, b = 445, and c = 600; ' 

find A = 63° 26' 09", B = 44° 47' 10", C = 71° 46' 36". 

12. a = 409 yds., b = 169 yds., and c = 510 yds; 

find the area of the triangle = 30,600 square yards. 

13. Given A' = 53° 24', B = 66° 27', and c = 338.65 yards; 

find a = 313.46 yards, and b = 357.92 yards. 

14. Given b = 483.74 feet, c = 379.41 feet, and B = 34° 11'; 

find A = 119° 40' 16", C = 26° 8' 44", and a = 748.12. 

15. Given b = 379.41, c = 483.74, and B = 34° 11'; 

find C = 45° 45' 10", a = 664.90, and a' = 135.43. 

16. Given B = 40°, b = 140.5 meters, and a = 170.6 meters; 

find A = 51° 18' 22", C = 88° 41' 38", and c = 218.55 meters. 

17. Given a = 374.5, b = 576.2, and c = 759.3 feet; 

find A = 28° 35' 40", and B = 47° 25' 22". 



50 PLANE TRIGONOMETRY 

18. Given A = 86° 19', b = 4930 feet, and c = 5471 feet; 

find C = 50° 1', B = 43° 40', and a = 7125.43 feet. 

19. Given a = 105.25 feet, b = 76.75 feet, and A' - B = 17° 48'; 

find A = 53° 54', B - 36° 6', C = 90°, and c = 130.26 feet. 

20. If b : c : : 4 : 5, a = 1000 yards, and A = 37° 19'; 

find B = 53° 37' 41", C = 89° 33' 19", and 6 = 1328.2 yards. 

21. Given c = 23.647 ch., a = 14.135 ch., and A = 33° 17' 18"; 

find B = 80° 2' 42", C = 66° 40', and b = 25.366, 
and B' = 33° 22' 42", C" = 113° 20', and b' = 14.169. 

22. Given a = 101.47 feet, c = 99.367 feet, and B = 47° 48' 12"; 

find A = 67° 27' 7", 5 = 64° 44' 41", and b = 81.40 feet. 

23. Given a = 363.24 feet, b = 146.18 feet, and C = 68° 14' 24"; 

find A = 88° 2' 36", 5. = 23° 43', and c = 337.6 feet. 

24. Given a = 115.03 feet, c = 112.06 feet, and b = 129.15 feet; 

find I A = 28° 12' 54", J 5 = 34° 39' 14", and } C = 27° 7' 52". 

25. Given a = 46.703 ch., 6 = 57.147 ch., and A = 19° 17' 42"; 

find B = 23° 50' 56", C = 136° 51' 22", and c = 96.654 ch., 
also B' = 156° 9' 4", C = 4° 33' 16", and c' = 11.223 ch. 

26. Given B = 38° 19' 25", c = 472.85, b = 294.84; (see § 41) 

find C = 83° 59' 15", C = 96° 00' 45", a = 401.87, a' = 340.09. 

27. Given a = 25.69 ch., b = 49.00 ch., and c = 50.35 ch.; 

find the area of the triangle = 615.52 sq. ch. 

28. Given A = 47°, B = 53°, and c = 263.57 yards; 

find the area of the triangle = 20,602 square yards. 

29. Given A = 37°, b = 161.79, and c = 149.91 yards; 

find the area of the triangle = 7298.0 square yards. 

30. One side of a parallelogram is 56 feet, and the angles between this 
side and the diagonals are 31° 14' and 45° 37'. Find the other side = 
44.95 feet. 

31. From the top of a bluff, the angles of depression of two points in 
the plane below, in line with the observer and 1000 feet apart, are found 
to be 27° 40' and 9° 33' respectively. Find the height of the bluff = 
247.74 feet. 

32. Two sides of a triangle are 5623 and 4977 feet, and the difference 
of the angles opposite to these sides is 15° 48' 32". Solve the triangle. 

Ans. 74° 12' 21", 58° 23' 49", side = 4301 feet. 

33. Two sides of a parallelogram are 65 and 133 yards, and one of the 
diagonals 159 yards. Find the angles of the parallelogram and the other 
diagonal. Ans. One angle = 101° 13' 47", diagonal = 136.19 yards. 



SOLUTION OF OBLIQUE TRIANGLES 51 

34. To find the distance of an inaccessible object A from B, a line 
BC = 208.3 yards, and the angles ABC = 126° 35' and ACB = 31° 48' 
were measured. Find AB. AB = 297.95 yards. 

35. To find the distance between two points A and B, a base line CD 
of 150 yards was measured. At C the angles ACD = 95° and BCD = 
70° were measured, and at D the angles BDC = 83° and A DC = 30° 
were measured. Find AB = 248.0 yards. 

36. A and B are two points on opposite sides of a mountain, C is a 
point visible from A and B; AC = 10 miles, and BC = 8 miles, and the 
angle BCA = 60°. Find AB = 9.165 miles. 

37. Two straight railways intersect at an angle of 60°. From their 
point of intersection two trains start at the same time, one on each line, 
and one at the rate of 40 miles per hour. Find the rate of the second 
train that at the end of an hour they may be 35 miles apart. 

Ans. 25 miles or 15 miles. 

38. From two stations A and B on shore, 3742 yards apart, a ship C 
is observed at sea. The angles BAC = 72° 34' and ABC = 81° 41' are 
measured. Find the distance to the ship = 8522.6 yards. 

39. The distance from N. to S. is 7282.66 miles, the angles JNS = 55° 
10' 55" and JSN = 83° 31' 52" were measured. Also on the other side 
of NS, KNS = 75° 16' 49" and KSN = 45° 55' 45" were measured. 
Find the distance JK = 156,459 miles. 

40. The parallel sides of a trapezoid are 628 and 417 yards. The 
angles at one end are 62° 13' and 117° 47', and at the other 84° 38' and 
95° 22', the non-parallel sides forming in each case acute angles with the 
longer of the parallel sides. Find the lengths of the non-parallel sides. 

Ans. 384.17 yards and 341.38 yards. 

41. The diagonals of a parallelogram are 102.12 yards and 141.16 
yards, and their included angle 72° 12'. Find the sides and angles. 

Ans. 98.952 yards, 73.385 yards, and 1^° q^ Qg'/ 

42. Two points M and N are inaccessible. A line AB = 350 yards is 
measured. At A the angles BAM = 102° 19' 18" and BAN = 41° 7' 12" 
are measured. At B the angles ABN = 98° 16' 24" and ABM = 
52° 17' 48" are measured. Find MN = 607.7 yards. 

43. In a triangle A = 46°, b = 121 feet, and c = 157 feet. The side 
a is divided into three equal parts, and a line is drawn to the point of 
division that is nearest to B. Find the length of this line. 

Ans. 135.8 feet. 



52 PLANE TRIGONOMETRY 

44. To find the distance between two inaccessible points A and B, a 
line 600 feet is measured so that DC cuts AB. The angles DCB = 58° 12', 
CDB = 49° 38', ACD = 74° 16', and ADC = 62° 13'. Find AB = 
1151 feet. 

45. A triangle with a = 27, 6 = 39, and c = 32, has a line drawn 
through B bisecting the area of the triangle. Find the parts into which 
B is divided. Ans. 45° 22' 36", 36° 54' 29". 

46. Three points A, B, and C are in the same straight line. AB = 1000 
feet, BC = 2000 feet. At a point D the angles ABD = BDC = 35°. 
Find AD = 1574.2 feet. 

47. The three medial lines of a triangle are 13, 19, and 17, — 13 to c, 
19 to b, and 17 to a. [See Geom. a 3 + 6 2 = \ c 2 + 2 (13) 2 .] Find a = 
18.51, b = 15.71, c = 22.42. 

48. A flagstaff 25 feet high stands on top of a cliff. From a point on 
the shore the angles of elevation to its top and bottom are observed to 
be 47° 12' and 45° 13'. Find the height of cliff = 348.3 feet. 

49. The angle of elevation of a balloon from a point due south of it is 
60°, and from another due west of the first and distant one mile it is 45°. 
Find the height of the balloon = 6466.6 feet. 

50. A, B, and C are three points distant AB = 1056 yards, AC = 924 
yards, and BC = 1716 yards. An observer places himself at P from 
which C appears directly in front of A, and measures the angle CPB — 
14° 24'. Find CP = 2110 yards. 

51. Find the area of a triangular field one of whose sides is 45 rods 
long and the two adjacent angles are 70° and 69° 40'. 

Ans. 8 acres, 98.4 poles. 

52. The area of a triangle is 84 square feet, and two of its sides are 
15 feet and 13 feet. (See eq. 95.) Find the third side = 14 feet. • 

53. The perimeter of a triangle is 400 feet, and its angles are 65° 15', 
75° 30', and 39° 15'. Find the sides. 

(a + b = 400 - c = C -^^ + C -^4 • Ans. 144.78, 154.35, 100.87. 

sin C sin C 

54. Find the perpendicular let fall from the vertices of a triangle upon 
the opposite sides when a = 25, b = 30, and c = 35. 

Ans. To c = 20.995. 

55. M and N are two inaccessible objects. A straight line AB = 280 
yards is measured. At A the angles MA B = 95° and NAB = 47° 30' 
are measured. At B the angles MBA = 52° 20' and NBA = 110° are 
measured. Find MN = 509.87 yards. 



SOLUTION OF OBLIQUE TRIANGLES 



53 



56. In a quadrilateral are given the four sides, a = 25.63, b = 24.09, 
c = 9.92, d = 29.97, and the angle, 78° 25', which the sides a and b 
make with each other. Find the angles and area of the quadrilateral. 

Ans. 125° 17' 53", 66° 59' 59", 78° 25', 89° 17' 08", area = 451.08. 

57. Given the two diagonals, 5 yards and 6 yards, of a parallelogram, 
and the angle 49° 18' which they form; find the sides and angles of the 
parallelogram. Ans. a = 2.339 yards, b = 5.004 yards, A = 76° 24' 20". 

58. Suppose the center of the earth at 
E and that of the moon at M, and let the 
circle, Fig. 13, represent the earth. At 
the stations S and S' the angles ZSM = 
44° 54' 21" and Z'S'M = 48° 42' 57" are 
measured. The angle SES' = 92° 14', and 
the radii of the earth ES and ES' = 3960 
miles. Find EM = 238,200 miles. 

59. The circles M and O, Fig. 14, 
represent the moon and earth. OE = 3960 
miles is the earth's radius, and M N that 
of the moon is to be found. The angle 
EMO = 57', and the angle NOM = 31' 
20". The angles MEO and MNO are right angles. Find the distance 
MO of the moon from the earth, and 2 MN the diameter of the moon. 

Ans. MO = 238,847 miles, MN = 2177 miles. 




Fig. 13. 




Fig. 14. 

The preceding examples have been solved by five-place log- 
arithmic or natural functions, so if six-place tables are used the 
results will be slightly different. The reason of this is due to 
the unavoidable errors in their last figures, as will appear in com- 
puting the logarithms of numbers to five places, beginning with 
100, by the formula 

log (* + 1) - log z + £2L . 

The computation should extend to six decimals, and the nearest 
fifth decimal only should be retained. 



CHAPTER V 

AREAS 

45. To find the radius R of a circle circumscribing a triangle. 
Draw the diameter CD equal to 2 R, Fig. 15. The angles 
D and A are equal, and DBC is a right triangle. 

Sin D = 2R = Sin ^' 



.*. 2R 



sin A sin B sin C 
By (93), k = J 6c sin A. Substituting sin A from (97) 



4 22' 4/c 



(97) 



. . . . (98) 




Fig. 15. 



Fig. 16. 



46. To find the radius r of a circle inscribed in a triangle. 
The triangles AOB, AOC, BOC compose the triangle ABC, Fig. 
16, and have for their bases a, b, c of a common altitude r; hence 

k = \ (a + b + c) r = rs. 



k Vs (s — a) (s — b) (s — c) 
r = _ = i '-± '-A u or 

s s 



r = 



v- 



(s - a) (s - b) d 



r) 



(99) 



This is the auxiliary quantity assumed in § 43. 

54 



AREAS 



55 



47. To find the angles and area of a quadrilateral inscribed in a 
circle. 




Let k equal the area, and 2s = a + b + c + d. 

D = 180° - B, .'. cos D = - cos B. 
By (89) and (61), AC 2 = a 2 + b 2 - 2 ab cos B, or 
AC 2 = (a - b) 2 + 4 a& sin 2 } 5. 
Also AC 2 = c 2 + d 2 + 2 cd cos B = (c + d) 2 - 4 cd sin 2 £ B. 
Equating and finding sin J B = J^tA) 2 



(ab + cd) 



> or 



sin § 5 = 



-v/ 



6 + c + d)(6 + c + 



a) 



4 (a& + cd) 



(s — a) (s — b) 
ab + cd 



(100) 



Likewise cos £ 5 



tan \ B 



-yl- 



(s — c) (s — d) 
ab + cd 



and 



V s - 



a) (s - b) 



(s — c) (s — d) • 
By (93) and (59), 

k = \ (ab + cd) sin 5 = (a& + cd) sin \ B cos £ B, 
or substituting, 



fc = V(s - a) (s - 6) (s - c) (s-d). 



(101) 



56 PLANE TRIGONOMETRY 

Solution of Trigonometric Equations 

48. Find x from sin (<f> ± x) = m sin x. 

By (15), and factoring, sin (<j> ± x) = sin <£ sin x (cot x ± cot (f>), 

whence by equating and reducing, cot x = — =F cot 6. 

sm<f> 

49. Find x from tan (</> + x) = m tan x. 

Divide by tan x, take by composition and division, and make 
m = tan 6. 

tan (0 4- x) 4- tan a; = tan Q + 1 
tan (^ + x) — tan x tan 6 — 1 

By (34) and (32) and tan 45° = 1, 

sin (0 + 2 x) = cot (0 - 45°) sin <£. 

50. Find x from tan (<j> + x) tan x = m. 
By composition and division, 

1 4- tan (4> + x) tan x _ m + 1 
1 — tan (</> + x) tan x wi — 1 

whence by (35), m = tan d and tan 45° = 1, 

cos (0 4- 2 x) = tan (45 — 6) cos <f>. 

51. Find x from sin (<£ ± x) sin x = m. 

By (23), cos <p - cos (<j> ± 2 x) = ± 2 sin (<£ ± x) sin x = ±2 m, 
whence cos (<£ ± 2x) = cos <j> =F 2 m. 

sin x = m. 



52. Find r and * from . r cog ^ = „ 



r; 



??2 

By dividing, tan x = - . 

By squaring, adding, and extracting the square root of the 
sum, 

a /~ t~ ; — i ^ n 

r = v wr + vr = - — = 

sin x cos x * 

53. Find x from m cos x 4- n sin x = <?. 

Assume r sin <£ = m, and r cos <£ = n, whence tan <j> = — , and 



r = Vra 2 4- n 2 , which give the values of the assumed r and <£. 



TRIGONOMETRIC EQUATIONS 57 

Then r sin cj> cos x + r cos <j> sin x = 5, or sin (^ + x) = - 



v m 2 + n 2 

-t?- j j £ ( r sin (a + x) = m. 

54. Find r and x from \ . ) Q / 

( r sin (3 + x) = n. 

By (19) and (20), 2 r sin [J (a + /9) + x] cos J (a - /?) = m + n, 
and 2 r cos [i (a + /?) + x] sin J (a + p) = m — n, 

whence tan [J (a + 8) + x] = ^-^- n tan 4 (a + 3). 

m — n 

Computation of Trigonometric Functions 

55. The development of sin x in terms x expressed in circular 
units is easier made by the Calculus, and is therefore assumed 
here. 

sin x = x - — + — — + — etc. . . (102) 

6 120 5040 362880 

The unit of x in this being the radian, the angle given in de- 
grees must be multiplied by— ^— = , § 5, which will 

6 F y 180° 57°.29578 ? 

introduce additional constant factors in the coefficients of the 
terms in the series. 

The computation is easier made by logarithms, therefore 
finding log x = log x° - log (57.29578) = 8.2418774 + log x° and 
the logarithms of the resulting coefficients, and using the nota- 
tion of § 29, log- 1 u for a number whose logarithm is u, the 
result is 

f log- 1 (8.2418774 + log x°) -log- 1 (3.9474809 + 3 . logx°)l 

sin x = \ +log- 1 (9.130206 + 5 . log x°) - log- 1 (3.9907 + 7 . logs ) \ 

t + log- 1 (8.617 + 9 . log x°) -etc. J 

. . . (103) 

Having found the natural sine in this way all the other func- 
tions may be determined from it (11). When all are computed 
thus from 0° to 45°, then sin x° = cos (90° - x°), cos x° = sin 
(90° — x°), etc., will give the functions from 45° to 90° without 
additional work. 



58 



PLANE TRIGONOMETRY 



As an example for illustration and one which will test the 
accuracy of the formula at the same time, find the sine of 45°. 



sin 45° = J 



log. of consts. =8.2418774 

log. powers of 45 =1.653212 5 

logs, of terms = 9. 8950899 



3.9474809 
4.9596375 



9.130206 
8.266062 



7.396268 



3.9907 
1.5725 



5.5632 



8.617 
4.879 



.00249040 



sin 45° = .78788889 



8.9071184 
cor. nos. = +.78539818 - .08074550 

- .00003658 + .00000031, 
08078208 = .7071068, true to 7 places. 



3.496 



Angles in degrees, minutes, and seconds must be reduced to 
degrees and decimals of a degree, as 27° 32' 57".32 = 27°.549256. 



sin 27°.549256 



log. of consts. =8.2418774 

log. powers of ang. = 1 4401099 

logs, of terms = 9. 6819873 



3.9474809 
4.3203296 



+ 



9.130206 
7.200549 



6.330755 



3.9907 
0.0808 



4.0715 



8.617 
2.961 



sin 27° 32' 57".32 



8.2678105 
corr. nos.= +.48082527 - .01852723 + .00021417 
- .00000118 + .00000000, 
48103944 - .01852841 = .4625110, true to 7 places. 



1.578 



A five- or six-place table of logarithms will give a result true to 
5 or 6 decimals. 

cos 27° 32' 57.32" = ^/l - (.46251 10) 2 = .8866137. 

From these the other functions are easily found. 

In the formula above, the fifth column is zero for all angles less 
than 28°, and the fourth for all less than 12°. 

The method is independent; that is, it does not depend upon 
the computation of any preceding functions, as is the case with 
the methods usually given. 

De Moivre's Theorem 



56. (cos x + V— 1 sin x) 2 = cos 2 x — sin 2 x 
+ \/— 1 (sin x cos x + cos x sin x), 

or (cos x + V- 1 sin x) 2 = cos 2 x + V— 1 sin 2 x. 

Multiply both members of this equation by cos x + V—l sin x, 
giving 
(cos x + V— 1 sin x) 3 = cos 2 x cos x — sin 2 x sin x 



+ v — 1 (sin 2 x cos x + cos 2 x sin x) , 
or (cos x + V- 1 sin x) 3 = cos 3 x + V - 1 sin 3 x 



AREAS 59 

Hence (cos x 4- V 7 — 1 sin x) n = cos nx + V — 1 sin nx, (104) 
n positive and integral. 

Extracting the nth. root of the last equation and raising then to 
the mih. power, gives 

m 

(cos x 4- V— 1 sin a;) m = (cos nx 4- V— 1 sin nz) n . 

Substituting x for nx and — for £, gives 

n 

m 

(or / oc\ ™" / n 

cos - + V — 1 sin - = (cos x 4- V — 1 sin x) , or, 
n n) 

reducing as above, 



(cos x + V — 1 sin x) = cos — x 4- V — 1 sin — x. 

n n 

Taking the reciprocal of (104) gives 

1 



(cos x + \/— 1 sin x)~ n = 



cos nx + V— 1 sin nx 
= cos nx — V — 1 sin nx, or 
(cos x 4- V— 1 sin x)~ n = cos (— nx) 4- V— 1 sin (— nz). 

Hence (104) is true for all values of n, and is the required 
theorem. 



PART II 

CHAPTER VI 
SPHERICAL TRIGONOMETRY 

57. Spherical Trigonometry treats of the relations between the 
face and diedral angles of Triedral Angles, and their solutions 
by means of the spherical triangles forming their bases. 

Three different planes passed through the center of a sphere 
form eight triedral angles, with vertices at the center, and 
having for bases the parts of its surface inclosed by their inter- 
sections. 

It is more convenient to use the spherical triangle which forms 
the base of a triedral angle in the investigation of its parts and 
their mutual relations. 

The face angles of the triedral angle are measured by the 
arcs of great circles which make the sides of the corresponding 
spherical triangle, when the radius of the sphere on which it is 
formed is the unit radian, and its diedral angles, measured in the 
same way, are the angles at the vertices of this triangle. These 
measures are also made in degrees. 

In either case, the parts expressed in degrees or circular units, 
will be correct for the same triedral angle whatever may be the 
radius of the sphere on whose surface its base may be; which 
practically allows the radius to be unity in every case. 

The angles of the spherical triangle are represented by A, B, C, 
and the opposite sides by a, b, c. 

A proper conception of the parts of the spherical triangle and 
their relation to its corresponding triedral angle is very necessary. 

Trace out the relation of the triedral angle to the spherical 
triangle ABC forming its base, in Fig. 18. The three planes 
AOB, AOC, BOC passing through the center intersect the 
surface in the arcs of great circles, AB, AC, BC, forming the 
triedral angle — ABC, with the vertex 0, and base the spheri- 

60 



SPHERICAL TRIGONOMETRY 



61 



cal triangle ABC. The face angles AOB, BOC, and COA about 
the vertex are measured by a, b, c respectively; and the 
angles of the triangle, denoted by A, B, C, are the respective 




Fig. 18. 

diedral or edge angles AO, BO, CO, or the angles between the 
planes ABO and ACO, BAO and BCO, and CAO and CBO. 

A review of the parts of Solid Geometry pertaining to the 
sphere should now be made, to fix in mind the following proper- 
ties of the spherical triangle there proved : 

1. Each side must be an arc of a great circle. 

2. The sum of any two sides is greater than the third side. 

3. The sum of its sides is less than 360° or 2 a: 

4. The greater side is opposite the greater angle, and conversely. 

5. The sum of its angles is greater than 180° or n, and less than 540° 
or 3tt. 

b' 




Fig. 19. 

6. If A'B'C ', Fig. 19, is the polar triangle of ABC, then conversely, 
ABC is the polar triangle of A'B'C. 



62 



SPHERICAL TRIGONOMETRY 



7. The following relations also exist between polar triangles : — 
a! = 180° 



A, V = 180° 


- B, c' 


= 180 - C, 


a, B' = 180° 


- b, C 


= 180 - c. 



A' = 180 c 

8. Each angle is greater than the difference between 180° and the 
sum of the other two; that is,A> 180° - (B + C) or > (B + C) - 180°. 

58. Spherical triangles having one, two, or three right angles, 
are called respectively spherical right triangles, birectangular 
triangles, and trirectangular triangles. 

If one, two, or three sides of a triangle are quadrants, the 
triangle is called respectively, quadrantal, biquadrantal, and 
triquadrantal; and the latter is one eighth of the surface of the 
sphere. 

Each of the parts of a spherical triangle will be taken less than 
180°, though they might be anything between 0° and 360°. 



B 90° 







Fig. 20. 



59. Relations between the Trigonometric Functions of the sides 
and Angles of a Spherical Right Triangle. 

Let ABC in each diagram, Fig. 20, represent a right triangle, 
right-angled at C. In each figure, I, II, III, make BC", BC ', 
B'C, and B'C" quadrants by prolonging BC, BA, and CA if 
necessary. Then it is evident that in each, CC" = co . a, AC 
= co . c, AB' = co .b, B'C = co . B, and that the angle AB'C 
is measured by co . a, since B' is the pole of the arc CC"; also 
that in each the triangle AB'C is right-angled at C. 

Let be the center of the sphere on which the triangle in I, 



SPHERICAL TRIGONOMETRY 63 

Fig. 20, is situated, and join A, B, and C to 0. Draw the tan- 
gents CE and CD and prolong OA and OB to meet these in E 
and D, and join ED. Then CD = tan .a, CE = tan . 6, OD 
= sec . a, and 0# = sec . 6. C being a right angle, DCE is a 
right triangle, and 

DW = tan 2 . a + tan 2 . b. 

Also in the triangle DOE by (89), 

Z>£ 2 = sec 2 . a + sec 2 .6 — 2 sec . a X sec . b X cos . c. 

Hence, tan 2 . a + tan 2 . b = sec 2 . a + sec 2 . b 

— 2 sec . a X sec . b X cos . c. 

Substituting (8) for sec 2 . a = tan 2 . a + 1, and for sec 2 . b 
= tan 2 .6 + 1, and canceling, gives 

zero = 2 — 2 sec . a X sec . 6 X cos c, 

whence cos . c = cos . a cos .6 (105) 

Applying (105) to the corresponding parts of AB'C in I, or 
either of the same in II, or III, Fig. 20, the parts being homo- 
logous and of like dimensions to ABC in I, gives, 

sin 6 = sin B X sin c . . (106) 

Interchanging a and 6, sin a = sin A X sin c . . (107) 

Applying (107) to AB'C', cos B = cos 6 X sin A . . (108) 
Interchanging a and 6, cos A = cos a X sin B . . (109) 

From (108) and (109) find cos 6 = ^? and cos a = ^L^ 

sin A sin B ' 

and substitute in (105), giving 

cos c = cot A X cot B . . (110) 

Applying (110) to AB'C, sin 6 = cot A X tan a . . (Ill) 
Interchanging a and 6, sin a = cot B X tan 6 . . (112) 

Applying (112) to AB'C, cos B = tana X cot c . . (113) 
Interchanging a and 6, cos A = tan 6 X cot c . . (114) 
Since the above ten formulae are obtainable from any one of I, 

II, III, Fig. 20, they are true for all possible values of the parts 

of a right spherical triangle. 



64 SPHERICAL TRIGONOMETRY 

If the side AC be produced to some point A" beyond C, and 
A" be joined to B by the arc d , forming a right spherical triangle 
CA"B, then from (106), 

sin a = sin A" X sin d, 

which compared with (106) gives for the oblique triangle ABA" 

sin A sin d 



sin A " sin c 



(Law of sines.) (114J) 



Hence, The sines of the angles of any spherical triangle are to 
each other as the sines of their opposite sides. 

From equation (105) the above formulae and proposition 
have "been proved, and these will directly or indirectly determine 
every formula in spherical trigonometry. 

Each of the above ten formulae involves three of the five 
parts a, b, c, A, B. As there can be but ten combinations of 
five things, three in a set, it follows, that for every possible case 
that can arise, any one part of any right triangle may be deter- 
mined if two others are given, and these are therefore sufficient 
to solve every case of right triangles. 

An easy method of obtaining any one of the ten when needed 
is devised by arranging them in two groups after replacing cer- 
tain of them by their complements in order to effect the classi- 
fication, thus: 

First Group 

sin a = tan (90 — B) tan b. 

sin b = tan (90 — A) tan a. 
sin (90 - c) = tan (90 - A) tan (90 - B). 
sin (90 - A) = tan b tan (90 - c). 
sin (90 - B) = tan a tan (90 - c). 

Second Group 

sin a = cos (90 — A) cos (90 — c). 

sin b = cos (90 - B) cos (90 - c). 
sin (90 — c) = cos a cos b. 
sin (90 - A) = cos (90 - B) cos a 
sin (90 - B) = cos (90 -A) cos 6. 



SPHERICAL TRIGONOMETRY 65 

The pans a. b. 9\f - A, 90 G - B. and 90° - c. are called 
Circular Parts, and are the only forms which enter the groupings, 
the right angle C not being one of the parts, and a and b on either 
side of it are not regarded as separated by it. Calling each part 
in succession the Middle Part, and the two immediately next to 
it. one on either side, the Adjacent Parts, and the other two the 
Opposite Paris, it is evident from the above groupings that the 
two mnemonic rules oi Napier will give the ten formulae, viz.: 

First: The sine of the middle part is equal to the product of the 
tangents of the adjacent parts. 

Second: The sine of the middle part is equal to the product oj the 
cosines of the opposite parts. 

These two rules applied to the five circular 
pans of either of the two triangles ABC or Co -%. 
AB'C in any one of figures 20 will give the 
ten formula?. Keep Fig. 21 in mind to aid in 
remembering the circular parts, and until per- 
fectly familiar with the application of the rules. 
It is well to draw this figure in the solution 
of every example. The right angle is at the 
vertex without a letter. To illustrate the mean- 
ings of adjacent and opposite parts, if co . A 
is a middle part co . c and b are adjacent parts, and co . B and 
a are opposite pans, each of the last two being separated from 
co . A by an intervening pan. The right angle not being one 
of the pans, the sides a and b are adjacent. To aid in remem- 
bering the rules, observe that o is in cosine and opposite, while 
a is in tangent and adjacent. 

It is a good exercise to determine all the formula? by applying 
the rules to Fig. 21. 

60. In the solution of spherical right triangles there are certain 
limitations existing between the pans which must be observed. 

From 'luS> and 109) are obtained 

■ cos B j , „ cos A 

sin A = . and sin B = 




cos b cos a 

A and B being less than ISO the first members are positive. 
and so the terms of the ratios in the second members must have 
the same sism: that is. both must be — or both — . Hence. 



66 SPHERICAL, TRIGONOMETRY 

An oblique ' angle and its opposite side in a right triangle must 
both be less or both greater than 90°, that is, must be in the same 
quadrant. 

So if a side and its opposite angle are given, there must always 
be two solutions. See Fig. 23. 

By (105), cos c = cos a X cos b. Hence, 

// c is less than 90°, a and b must both be less or both greater than 
90°, in order that the second member shall have the positive sign; if 
c is greater than 90°, a and b must be in different quadrants that the 
second member may be negative. 

From (107) and (106), sin A = ^-^- , and sin B = -^A m 

sin c sin c 

Since the sin A and sin B cannot be greater than 1, the sin a 
and the sin b must be equal to or less than sin c. Hence, 

c must always be nearer 90° than either a or b. Also if a 
or b is obtained from one of these equations, the sine being the same 
for its supplemental value, the true value must be determined by the 
principle, viz., that a and A or b and B are in the same quadrant. 

A and B denoting the oblique angles of any spherical right tri- 
angle, by (16) and (17), 

cos (A ± B) = cos A . cos B =F sin A . sin B, or 

cos (A ± B) =GO tA.cotBTl =cosc =F 1 by (110). Hence, 
sin A sin B 

The sin A . sin B being always positive, the sign of cos (A + B) 
must be the same as cos c — 1, that is, always negative, and the 
sign of cos (A — B) must be the same as cos c + 1, always positive. 

Therefore, A + B > 90 and < 270, and A - B < 90. 

Summary of Results in this Paragraph 

First. A and a, also B and b, in the same quadrant. 

Second, c < 90, a and b in same quadrant, c > 90 a and b in 
different quadrants. 

Third, c always nearer 90° than a or b, their values deter- 
mined by first. 

Fourth. A + B > 90 and < 270, A - B < 90. 



SPHERICAL TRIGONOMETRY bi 

Illustration of the 3Iethod of Solving any Case 

61. The other parts of any spherical right triangle can be 
found from any two given circular parts, as has already been 
shown. 

Suppose the given parts are a and c, and draw Fig. 21. 
To find B. In the figure, B is between a and c; then by rule 
first, 

sin (co . B) = tan a . tan (co . c), or cos B = tan a cot c. 

To find A. Of the given parts a and c and the required A, 
co . A and co . c are together, but each is separated from a by an 
intervening part, making a the middle part and co . A and co . c 
the opposite parts. Then by rule second, sin a = cos (co . J.) cos 
(co . c) , or sin a = sin A . sin c. 

To find b. Of the given parts a and c and the required 6, co . c 
is the middle part and a and b are opposite, and by second rule, 

sin (co . c) = cos a . cos b, or cos c = cos a X cos b . 

The solution may now be completed by the formulae thus 
determined, 

cos B = tan a . cot c, 

a sin a 
sin A = — — , 
sin c 

7 cos c 

cos o= . 

cos a 

In a similar method the required formulae for any of the six 
cases may be found by applying the two rules to Fig. 21. 

The required should always be found from the given parts, 
which can be done, since the two given and the one required will 
form a group of either a middle and two adjacent, or two oppo- 
site parts. If a computed should be used to get a required part, 
the error in the result due to the tables may be nearly doubled 
thereby. 

62. A Quadrantal Triangle can always be solved by its polar, 
which is a right triangle, or by applying Napier's rules to the 



68 SPHERICAL TRIGONOMETRY 

right triangle ADC, Fig. 22, formed by making BD a quadrant 
and drawing a great circle arc AD. The following relations of 
parts, however, must be observed : 

First. If the sides a and b or the angles A and B are in the 
same quadrant, C is greater than 90. See (117) and (120). 

Second. A and a, also B and b, are always in the same quad- 
rant. See (115) and (116). 

To solve the Isosceles Triangle, draw an arc of a great circle 
bisecting its vertical angle and base, then solve either right tri- 
angle, so formed, and the required parts of the isosceles triangle 
are determined. 




EXAMPLES AND EXERCISES IN RIGHT TRIANGLES 

It is customary to divide the solution of right triangles into six cases, 
the given parts of which are as follow : 

A side and hypotenuse. An angle and a side opposite. 

Hypotenuse and adjacent angle. The sides about the right angle. 
A side and an adjacent angle. The two oblique angles. 

When the solution is made by Napier's rules this classification is of 
no service, and is therefore ignored. 

1. Given c = 140° and a = 20°; solve the triangle. 

Applying the rules to Fig. 21 as above, sin (co . B) — tan a . tan (co . c), 
sin (co . c) = cos a . cos b, and sin a = cos (co . A) cos (co . c) } or, after 
reducing, 



SPHERICAL TRIGONOMETRY 69 



r> , <. a sin a 7 cos c 

cos B = tan a . cot c, sin A = cos 6 = . 

sin c cos a 

log tan a = 9.56107 log sin a = 9.53405 log cos c = 9.88425 n * 

log cote =0.07619"* cologsinc =0.19193 colog cos a = 0.02701 



log cos B = 9.63726 n * log sin A =9.72598 log cos b = 9.91 126 n * 

180° - B = 64° 17' 35", 180° - b = 35° 23' 40" 

B = 115° 42' 25", A = 32° 8' 48", b = 144° 36' 20". 

2. Given c =110° 46' 20" and A = 80° 10' 30"; solve the triangle, 
sin (co . c) = tan (co . A) tan (co . B), sin (co . .4) = tan b . tan (co . c), 

and sin a = cos (co . A) cos (co . c), or reducing, 

cot B = cos c . tan A, tan 6 = cos A . tan c, sin a = sin A . sin c. 

log cos c = 9.54980 n * log tan c = 0.42100 n * log sin A = 9.99358 
log tan A =0.76150 log cos A = 9.23208 log sin c =9.97081 

log cot B = 0.31 130 n * log tan b = 9.65308 n * log sin a = 9.96439 
180° - B = 26° V 38", 180° - b = 24° 13 r 16". 

B = 153° 58 r 22", b = 155° 46' 44", a = 67° 6' 48". 

3. Given a = 31° 20 r 45" and B = 55° 30 r 30"; solve the triangle. 

sin (co. B) = tan a tan (co . c), sin (co . A) = cos a cos (co . B), sin a — 
tan 6 . tan (co . B), after reducing 

cot c = cos B . cot a, cos A = cos a . sin B, tan 6 = sin a . tan 5. 

log cos B = 9.75304 log cos a = 9.93148 log sin a = 9.71617 
log cot a = 0.21531 log sin B = 9.91604 log tan B = 0.16300 



log cot c = 9.96835 log cos A = 9.84752 log tan b = 9.87917 
c = 47° 5 r 9", A = 45° 15 r 28", b = 37° 7' 49 r 




Fig. 23. 

4. Given A = 76° 12 r 12" and a = 37° 36 r 24"; solve the triangle. 
By § 60 there are two solutions. The figure shows that each triangle 
contains the given parts, and the two together make up the lune AA'. 

* n above and to the right of a logarithm mean 6 - A he number correspond- 
ing is negative. 



70 SPHERICAL TRIGONOMETRY 

sin b = tan a . tan (co . A), sin a = cos (co . A) cos (co .c), sin (co . A -- 
cos a cos (co . B ), or sin 6 = tan a . cot A, 

sin a = sin A . sin c, cos A = cos a . sin B, 

sin a . D cos A 

. . sin c = — — 7 > ■ sm ij = • 

sin A cos a 

log tan a - 9.88665 log sin a = 9.78550 log cos A = 9.37745 
log cot A = 9.39018 colog sin A = 0.01271 colog cos a = 0.10116 

log sin b = 9.27683 log sin c = 9.79821 log sin B = 9.47861 

b = 10° 54' 14", c = 38° 55' 45", B = 17° 31' 10". 

V = 169° 5' 46", c f = 141° 4 r 15", B = 162° 28' 50' 



6. Given a = 132° 6' and 6 = 77° 51'; solve the triangle, 
sin (co . c) = cos a . cos 6, sin b = tan (co . A) tan a, 
sin a = tan (co . B) tan o, or 

, , , tan a , u tan 6 

cos c = cos a cos o, tan A = — — — , tan B = —. . 

sin o sin a 

log cos a = 9.82635 n log tan a = 0.04404 71 log tan b = 0.66697 
log cos b = 9.32319 colog sin b = 0.00984 colog sin a = 0.12961 

log cos c = 9.14954 n log tan A = 0.05384 n log tan B = 0.79658 
180° - c = 81° 53' 18", 180° - A = 48° 32' 32", B = 80° 55' 27'> 

c = 98° 6' 42", A = 131° 27' 28". 

6. Given A = 91° 11' and B = 111° 11'; solve the triangle. 

sin (co . c) = tan (co . A) tan (co . B), sin (co . A) = cos (co . 2?) cos a, 
sin (co . B) = cos (co . A) cos 6, whence 

i a i. r> cos A , COS B 

cos c = cot A . cot B, cos a = - — - , cos b = - — - . 

sin B sin A 

log cot A = 8.31505 log cos A = 8.31495" log cos B = 9.55793" 
log cot B = 9.58832« colog sin B = 0.03038 colog sin A = 0.00009 

log cos c = 7.90337 log cos a = 8.34533 n log cos b = 9.55802™ 

180° - a = 86° 43' 52", 180° - b = 68° 48' 44". 
c = 89° 32' 28", a = 91° 16' 8", b = 111° 11' 16'. 



SPHERICAL TRIGONOMETRY 71 

7. c = 90, a = 67° 38', and b = 48° 50' ; solve the triangle. 
This is a quadrantal triangle, and by reference to Fig. 22 and "rela- 
tions 1st and 2d" given with it the required formulae are: 

Given a and b, Given A and B, 

— cos C = cot a cot b, — cos C = cos A cos B, 

cos 6 = cos B sin a, sin A = tan 5 cot 6, 

cos a = cos A sin 6. sin B = tan A cot a. 

log cot 6 = 9.94171 log cos 6 = 9.81839 log cos a = 9.58039 

log cot a = 9.61436 colog sin a = 0.03397 colog sin b = 0.12332 



log cos C = 9.55607 n log cos B = 9.85236' log cos A' = 9.70371 
180° - C = 68° 54' 42", B = 44° 37' 5", A = 59° 38' 11' 

C = 111° 5' 18". 

8. Given a = 40° 42' 24" and c = 63° 20'; 

find b = 53° 41' 56", A = 46° 52' 15", B = 64° 4' 2". 

9. Given a = 70° 15' 30" and A = 81° 42' 42"; 

find b = 23° 56' 58", B = 25° 20' 19", c = 72° 1' 0", and 
6' = 156° 3' 2", B' = 154° 39' 41". 

10. Given b = 30° 32' 24" and A' = 36° 44'; 

find a = 20° 46' 1", c = 36° 21' 35", B = 58° 59' 40". 

11. Given a = 28° 47' and b =110° 27' 18"; 

find c = 107° 50' 11", A = 30° 23' 4", B = 100° 10' 53". 

12. Given c = 72° 10' and A = 30° 43'; 

find a = 29° 5' 37", b = 69° 29' 2", 5 = 79° 41' 15". 

13. Given A = 40° 42' 24" and B = 67° 51' 36"; 

find a = 35° 4' 27", b = 54° 42', c = 61° 46' 35" 

14. Given c = 118° 49' and 6 = 44° 23'; 

find A = 122° 34' 33", a = 132° 24' 39", B = 52° 58' 7". 

15. Given A = 102° 57' and B = 143° 46'; 

find a =112° 16' 52", b = 145° 51' 37", c = 71° 42' 43". 

16. Given a = 10° 18' and 6=7° 10'; 

find A = 55° 31' 51", B = 35° 6' 58", c = 12° 32'. 

17. Given c = 59° 3' and A = 147° 32'; 

find a = 152° 35' 04", B = 108° 7' 9", 6 = 125° 24' 13". 

18. Given A = 98° 34' and a = 113° 12'; 

find B = 22° 13' 4", b = 20° 34' 39", c = 111° 38' 36", and 
B' = 6' = c' = 



72 SPHERICAL TRIGONOMETRY 

19. Given B = 100° 40' and a = 170° 38'; 

find A = 165° 50' 20", b = 139° 10' 10", c = 41° 42' 25". 

20. Given c = 90°, C = 98° 22' 42", and A = 150° 47'; 

find a = 150° 26' 14", b = 94° 43' 29", £ = 99° 36' 30". 

21. Given c = 90°, a = 121° 30' and B = 112° 16' 12"; 

find b = 108° 51' 8", 4 = 123° 30' 45", C = 102° 4' 40". 

22. Given c = 90°, a = 94° 22' 12", A' = 108° 13' 18"; 

find b = 14° 6' 14", B = 13° 25' 20", C = 72° 17' 30", 
and V - B'" - C" - 

23. Given A' = £ = 27° 12' and c = 135° 21'; 

find a = 69° 56' 10", C = 160° 00' 6". 

24. Given a = 6 = 112° 25' and c = 123° 48'; 

find A = 140° 35' 5", C = 145° 11' 30". 

25. Given a = b = 152° 6' and C = 67° 46'; 

find A = 120° 41' 21", c = 30° 14' 38". 

26. If A = 36° and £ = 60°, a + b + c = 90°. 

27. If A = a, show that the remaining parts of the spherical right 
triangle are each equal to 90°. 

28. If the sides of a spherical right triangle are equal to a, find the 
formulae for the angles. 

29. The observed altitude of two stars, one due north and the other 
due west, are respectively 52° 9' and 36° 37'; find the angular distance 
between them. Ans. 60° 29' 44". 

30. Prove in the right triangle sin 2 \c — sin 2 \ a cos 2 \ b + cos 2 \ a 
sin 2 \ b. See (109), (61), (60) 

31. Prove in the right triangle tan 2 \ A = S ? n ^ ~ ^ . (114) 

sm (c + b) 

32. Prove in the right triangle tan 2 \ b — tan \ (c + a) tan £ (c —a). 

(105), (4) 

33. If y be the length of a great circle arc drawn from the right angle 
perpendicular to the hypotenuse of a right triangle, prove 

cot 2 y = cot 2 a + cot 2 6, and cos 2 y = cos 2 A + cos 2 B. 

34. In the right triangle prove ^ = ^|4 ■ ( 108 )> ( 109 ) 

cos o sin z r> 

35. In the right triangle prove sin A sin 2 b = sin c sin 2 B. 

(106), (108) 

36. In the right triangle prove tan 2 % c C0S ^ ^ ^ . § 60, (110) 



SPHERICAL TRIGONOMETRY 



73 



37. In the right triangle prove tan 2 (45° - A) = ^-f + • 

(110), (61) 

38. In the right triangle prove sin (c - 6) =2 sin 2 £ A cos 6 sin c. 

(114) 

39. In the right triangle prove sin {c - a) = sin b . cos a . tan ^ 5. 

(106), (113), (61) 

40. If from any point in a trirectangular triangle arcs of great circles 
are drawn to the vertices, prove 

cos 2 a + cos 2 P + cos 2 7=1. 

Suggestion. Prolong AP to D and BP to E, PD = 90° - p, PE = 
90° - 7, and PDC and P#C are right triangles. 




The sum of the squares of the cosines of the angles which a straight line, 
PA, forms with three rectangular axes is equal to unity. 

AC, AB and a perpendicular to the paper at A are the axes. 

41. Two planes intersect at right angles along a line /. From any 
point R in I one line is drawn in each plane making respectively angles 
.4 and B with /. Find the angle between them = C. 

Prove cos C = cos A cos B. 




42. The sum of the angles of a splierlcal right triangle is less than 
four right angles. Why? See § 60. 



CHAPTER VII 



SPHERICAL TRIANGLES IN GENERAL 



63. Relations of the Trigonometric Functions of the Sides and 
Angles of Spherical Triangles in General. 

In Art. 59 it is proved that the formulae for the right triangle 
are true for all possible values of its parts; also the law of sines 
for all such values in the oblique triangle; therefore, as the form- 
ulae for the oblique are determined by means of those of the 
right triangle, Fig. 24 will suffice to illustrate the solutions of the 
different cases occurring in the various combinations of parts 
greater or less than 90, and also the determination of the form- 
ulae from (115) to (123). 







~-^ D 



ABC is an oblique triangle, BD and BE are made quadrants, 
FD and FC are not unless the triangle is right, which is not this 
case. CD and AE are complements of a and c, and ED meas- 
ures the angle B. The triangles AEF and CDF are evidently 
right triangles, and from them all the formulae of this paragraph 
are determined. The equations (a) to (h) are true for any tri- 
angle when A, C, a, c, and F are used in their proper quadrants. 

74 



SPHERICAL TRIANGLES IN GENERAL 75 

From the triangle AEF, 

cos c = sin F sin FA (a) 

cos A = sin F cos FE . . . . 8 . (b) 
sin c cos FE = cos FA. 

Multiplying last two and canceling, 

sin c cos A = sin F cos FA (c) 

By law of sines (114J), 

sin A cos c = sin F sin FE (d) 

It is evident that the corresponding equations for the triangle 
CDF are found from these by substituting a for c, 180 — C for 
A, FC for FA, and FD for FE, giving 

cos a = sin F sin FC (e) 

- cos C = sin F cos FD . . . . . . (f) 

— sin a cos C = sin F cos FC (g) 

sin C cos a = sin F sin FD ....... v (h) 

From (e), 

cos a = sin F sin (FA + b) = sin F sin FA cos 6 
+ sin F cos FA sin 6, 

and substituting from (a) and (c), 

cos a = cos b cos c + sin b sin c cos A . . (115) 
Interchange a and b, 

cos 6 = cos a cos c + sin a sin c cos B . . . (116) 
Interchange 6 and c, 

cos c = cos a cos b + sin a sin b cos C . . (117) 

Hence, The cosine of any side of a spherical triangle is equal to 
the product of the cosines of the other two sides increased by the 
product of their sines and the cosine of their included angle. 

From (b), 

cos A = sin F cos (FD — B) = sin F cos FD cos B 

+ sin F sin FD sin B, 

and substituting from (f) and (h), giving 

cos A = — cos B cos C + sin B sin C cos a . . (118) 



76 



SPHERICAL TRIGONOMETRY 



Interchange a and b, 

cos B = — cos A cos C + sin A sin C cos b . . (119) 
Interchange c and 6, 

cos C = — cos A cos i? + sin A sin 5 cos c . . (120) 

Hence, The cosine of any angle of a spherical triangle is equal to 
minus the product of the cosines of the other two angles increased by 
the product of their sines and the cosine of their included side. 

From (c), 

sin c cos A = sin F cos (FC — b) = sin F cos FC cos b + sin F 
sin FC sin b i 

and substituting from (e) and (g), 

sin c cos A = cos a sin b — sin a cos 6 cos C 
Interchange a and 6, 

sin c cos 5 = cos b sin a — sin b cos a cos C 
Interchange a and c, 

sin a cos 5 = cos b sin c — sin b cos c cos A 
Interchange b and c, f (121) 

sin a cos C = cos c sin 6 — sin c cos b cos A. 
Interchange a and b, 

sin 6 cos C = cos c sin a — sin c cos a cos 5. 
Interchange a and c, 

sin 6 cos A = cos a sin c — sin a cos c cos 5. , 

From (d), 

sin A cos c = sin F sin (FD — B) = sin F sin FD cos J5 
- sin F cos FD sin B, 

and substituting from (g) and (h) gives 

sin A cos c = sin 5 cos C + cos 5 sin C cos a. 
Interchange b and c, 

sin J. cos b = sin C cos 5 + cos C sin B cos a. 
Interchange a and 6, 

sin 5 cos a = sin C cos A + cos C sin A cos 6. 
Interchange a and c, 

sin B cos c = sin A cos C + cos A sin C cos b. 
Interchange b and c, 

sin C cos 6 = sin A cos B + cos A sin 5 cos c. 
Interchange a and b, 

sin C cos a = sin B cos A + cos B sin A cos c. 



(122) 



(123) 



GENERAL FORMULAE 77 

Divide the first of (121) by sin a, replace ^-^ by sin , trans- 

sm a sin A 

pose, and reduce, giving 

cot a sin 6 = cos b cos C + sin C cot A . 
Interchange b and c, 

cot a sin c = cos c cos Z? + sin B cot A . 
Interchange a and 6, 

cot b sin c = cos c cos A + sin A cot -B. 
Interchange a and c, 

cot 6 sin a = cos a cos C + sin C cot B. 
Interchange b and c, 

cot c sin a = cos a cos 5 + sin B cot C 
Interchange a and 6, 

cot c sin b = cos 6 cos A + sin A cot C. 



The groups (121), (122), (123), are not useful except in Geodesy 
and Astronomy. 

A good exercise for the student: Find all the formulae for 
right triangles by making C = 90° in equations (115) to (123). 



General Formulae Adapted to Logarithmic Computation 

64. Find by transposition and division from (115), 

, cos a — cos b cos c /-io/i\ 

cos A = .... (124) 

sin b sin c 

Q i , , • t -, -. A -. cos a — cos b cos c 

subtracting irom 1,1 — cos A = 1 — . 

sin b sin c 

Reducing by (61) and (17), 2 sin 2 U= — (6 ~ c) ~ C0Sa . 

sin b sin c 

Reducing by (22), 

. 2 x . _ 2 sin J (q + b - c) sin £ (a - 6 + c) 
2 sin 6 sin c 

Let 2 s = a + 6 + c, .*. a + 6 — c = 2(s — c),a — 6 + c 
= 2 (s - 6). 



78 SPHERICAL TRIGONOMETRY 



Substituting, sin i A = , / sin (s - b) sin (s - c) > 

V sin b sin c 

Likewise, sin \ 5 = . / sin (s - a) sin (g - c) < , 

V sin a sin c 

Likewise, sin \ C = i / sin ( s ~ a ) sin ( s ~ 5 ) . 

V sin a sin 6 

Adding 1 to both sides of (124), 

1 . A 1 , cos a — cos b cos c 

1 + cos A = 1 + : — ; — : 

sin 6 sin c 

By (62) and (16), 

2 cos 2 U= cos a - cos (b + c) < 

sin b sin c 

By (22), cos 2 J 4 = ^^j (a + b + c) sin j (b + c-a) 

2 sin 6 sin c 

Substituting as before, 



t A . /sin s. sin (s — a) 

cos £ y4 = 1/ ; — ~-i L 

y sin o sin c 



Likewise, cos }B - ./ sin «. sin (« - ft) ^ _ _ ( 

> sin a sin c 

t -i • i •> /sin s. sin (s — c\ 

Likewise, cos \ C . = \/ ; — - — ' • 

V sin a sin 6 

Dividing (125) by (126), 



tanii= y/? in 



sin s. sin (s — a) 



Likewise, tan J B = t/ gin (§ " "? " I" ( ' ~ C) " (127 > 

V sin s. sin (s — 6) 

Likewise, tan :* C = l/ sin ( f ~ a > si f ( I^. 

V sin s. sin (s — c) 

Again, from (118) is found 

cos a = cos A + cos B cos C (m) 

sin 5 sin C 
Subtracting from 1, 

cos A + cos B cos C 



1 — cos a = 1 



sin B sin C 



GENERAL FORMULAE 



79 



By (61) and (16), 



By (21), 



sm* 



— cos A — cos (B + C) 
sin B sin C 

i -2 cos (A+B + C) cos (B + C-A) 

2 sin 5 sin C 



Let2S = A + B + C .-. B + C- A = 2 OS - A), 

A + B - C = 2 (S - C), A + C - B = 2 (S - B). 



(129) 



Substituting, 


sin J a = 


^ 


- cos & cos (S— A) 
sin i? sin C 


Likewise, 


sin J 6 = 


y/= 


cos $ cos (*S — B) 
sin A sin C 



Likewise, sin J c = 

Adding 1 to both sides of (128), 



/- cos S cos QS - C) 
V sin A sin B 



By (62) and (17), 



1 + cos a 



2 cos 2 h a 



1 + 



cos A — cos B cos C 



sin B sin C 
cos A + cos (B — C) 



cos^ * a 



By (21), 
Substituting as before, 

cos \a = 

Likewise, cos \ b = y 



sin 5 sin C 
_ 2 cos i (A + fl-C) cos j (A-B + C) 



2 sin B sin C 



/ cos QS - E) cos QS - C) 

V sin i? sin C 

cos QS - A) cos QS - C) 
sin A sin C 



T ., . i 4 /cos OS - A) cos OS - 5) 

Likewise, cos * c = 1/ — : f—. — 

V sm A sm B 

Dividing (129) by (130), 

+ i / — cos S cos (S — A) 

tan ^ a " V cos QS -g) cos (S -C) ' 

/ - cos S cos OS - B) 

V I 



(130) 



Likewise, 



tan hb = 



Likewise, 



tan ^ c 



cos QS - A) cos QS - C) 
I - cos £ cos QS - C) 
; cos OS - A) cos OS -B) 



(131) 



80 SPHERICAL TIUCxONOMETKY 

Gauss's Equations and Napier's Analogies. 

65. Prolong the sides AC and AB of the triangle ABC to 
complete the lune AA' '. Make the constructions about the tri- 
angle with arcs of great circles oniy. Bisect the longer side AB 
at right angles by EM. Make ABM = A. Bisect MBC by BE, 
and join HC, which last will bisect BCM = 180° - C, thus 
making FCH = 90° — \ C, since H is equally distant from BM, 
BC, and MC. Draw HF perpendicular to BC, then CHF = EHB, 
each being supplements of BUM. (If CHF turned about H until 
HF falls on HE, HE will pass through A.) Make BD equal to 
AC, then FD will be equal to FC. 



It is now evident that CF = \ (a - b), FB = \ (a -f 6), ### 
= £ (A + 5), and H5i^ = J (A - B). 
From the right triangles CFH and ###, 

cos EHB = cos J c . sin | ( A + 5) , 
cos CHF = cos \ C . cos £ (a — 6). 

The first members being equal, 

cos J c . sin \ (A + 5) = cos £ C . cos \ (a - 6). 
Apply this to £CA/ and reduce, 

sin h c . cos % (A - B) = sin £ C . sin £ (a + 6). 
From 5CH the law of sines gives, 

sin \ (A - B) _ sin CH 
cos \ C sin HB ' 

and from CFH and EHB, 

smj (a-b ) = sin CH . sin Cff F = sin CH 
sin 4 c sin #5 . sin £H5 sin #5 ' 



GENERAL FORMULAE 81 

Hence equating the above values and reducing, 

sin ^ c . sin \ ( A — B) = cos | C sin \ {a — b) . 

Apply this last to the triangle BCA' and reduce, 

cos J c . cos \ {A + B) = sin J C . cos J (a + 6). 

These four grouped for convenient reference are: 

Gauss's Equations 

sin | c . sin | (A — 5) = cos J C . sin J (a - 6). . (132) 

cos \ c . sin i (A + B) = cos i C . cos \ (a - 6). . (133) 

sin £ c . cos i (A - B) = sin i C . sin J (a + 6). . (134) 

cos % c . cos J (A + 5) = sin J C . cos J (a + 6). . (135) 

These may be used in solving triangles when the three given 
parts are consecutive. 

Napier's analogies may be determined directly from Fig. 28, 
but more simply from Gauss's equations as follows: 

Napier's Analogies 

(132) - (133), tan J (a - b) = tan h ° ' sin * (A ~ B) • (136) 

sin 2 (A -f- z>) 

(134) H- (135), tan * (a + 5) - tan | C cos i (.4 - _g) . 

cos J (A + B) 

(132) -*-(134), tan J (A - 5) = cot * C ' sin * (a ~ & > ■ (138) 

sin ^ (a + b) 

(133) +(186), tani (.4 + 5) = «* *C • cos* (« -&) . (139) 

cos (a + 6) 



CHAPTER VIII 

SOLUTION OF OBLIQUE SPHERICAL TRIANGLES 

66. When a part of a triangle is obtained from its sine it may 
have either or both of two supplementary values, since the sine 
of an angle is equal to the sine of its supplement, in which case 
the proper value must be determined by other considerations, 
some of which are the following: 

cos A = cos a — cos b cos c 



By (115), 



sin b sin c 



whose denominator is always positive, and if a differs more from 
90° than b or c, its cosine is numerically greater, and therefore 

cos a > cos b cos c, 

because one of the factors cos b or cos c will be less than cos a, 
and the other cannot be greater than unity. The sign of the 
numerator and also of cos A will be the same as that of cos a, or 
a and A must be in the same quadrant. Hence, 

I. When a side, greater or less than 90°, differs more from 90° 
than another side, it is in the same quadrant as its opposite angle. 

By (118), cosa= cos A + cos B cos C , 

sin B sin C 

and as in I, if A differs more from 90° than B or C, it determines 
the algebraic sign of the fraction, so a and A must be in the same 
quadrant. Hence, 

II. When an angle, greater or less than 90, differs more from 90 
than another angle, it is in the same quadrant as its opposite side. 

As there will always be one side or angle differing more from 
90 than either of the other two, it follows that, 

III. In every spherical triangle there are two sides in the same 
quadrant as their opposite angles. 

82 



SOLUTION OF OBLIQUE SPHERICAL TRIANGLES 83 

From (137) , tan J (a + b) cos J (A + 5) = tan J c cos J (A — J?), 
the second member of which evidently must always be positive, 
and therefore the factors in the first must have the same signs. 
Hence, 

IV. The half sum of any two sides is in the same quadrant as the 
half sum of their opposite angles. 

67. Every case of oblique spherical triangles may be solved 
by Napier's rules applied to the two right triangles into which it 
may be divided by a perpendicular from a vertex judiciously 
chosen to its opposite side; but, excepting perhaps first and 
second cases, the methods by the formulae for oblique triangles 
are more simple. It is better to understand thoroughly one 
method in each case, and that the best, than several imperfectly, 
so the "perpendicular method" is not given. 

Since the solution requires three given parts, there can be but 

Six Cases 

I. Given two sides and their included angle; 
II. Given two angles and their included side; 

III. Given two sides and an angle opposite one of them ; 

IV. Given two angles and a side opposite one of them; 
V. Given the three sides; 

VI. Given the three angles. 

Case I. Given two sides and their included angle, a, b, C; solve 
the triangle. 

The required formula? are from (139), (136), and "Law of 
Sines." § 59. (114^) 

i / a , r>\ cos ^ (a — b) , i n 

tan \ (A + B) = J-J — —-f cot \ C. 

cos J (a + b) 

i / a d\ sin k (a — b) , , „ 
tan J (A - B) = - — =-± -( cot J C. 

sin § (a + 0) 

sin C sin b 

sm c = — — — • 

sin B 



84 SPHERICAL TRIGONOMETRY 

Example. Given a = 72°, b = 47°, C = 33°; solve the triangle. 

i (a + b) = 59° 30', \ (a - b) = 12° 30'. 

log cos J (a- 6) =9 .98958 log sin i (a -b) =9. 33538 log sin 0=9.73611, 
colog cos g (a+6) =0.29453 colog sin \ (a+6)=0 .06468 log sin 6=9.86413, 
log cot i C = 0.52840 log cot \ C =0.52840 colog sin 5=0 .18355, 
log tan \(A + B) =0.81251 log tan § {A - B) = 9 . 92846 log sin c= 9. 78739 
§(.! + £)= 81° 14' 45", § (^L -5) =40° 18' 7", c=37°26'0". 

^4 = 121° 32' 52", 5=40° 56' 38", 

since c and C same quad. 

Check by sin a sin Z? = sin b sin A \ ]°S sin B = ^qJ«o? !° § ^ 4 = 2' ££?£ 
J I log sin a= 9. 97821 log sin 6=9.86413 

9.79466 nearly = 9. 79467 

If only the side c is required, it may be found by natural func- 
tions from (117), cos c = cos a cos b + cos a sin b cos C, or, trans- 
forming by (44), 

cos r = cos a cos b + sin a sin b — 2 sin a sin 6 sin 2 J C 
= cos (a — b) — 2 sin a sin b sin 2 J C. 
Case I and Case II may also be solved by Gauss's Formulae. 
68. Case II. Given B, C, a; solve the triangle. 
From (137), (136), and " Law of Sines,' 7 (114£) 

tan i (c + b) = C ° S \ L 7 p tan i a > 
cos J (C + B) 

tan J (c - b) = sin i ( ^ ~ B) tan J a, 
2 ; sin i (C + B) 2 ' 

, . A sin B sin a /11/ln 

and sin A = : — - — (114^) 

sin o 

Example. Given 5 = 95° 38' 4", C = 97°26' 29", a = 64° 23' 15"; 
solve triangle. 

i (C + 5) =96 ° 32' 16", | (C - 5) = 54' 12". 

logcosi (C-5) =9.99995 log sin | (C- 5) =8 .19770 log sin 5=9.99790 

colog cos § (C+JB)=0.94364«cologsin£ (C + B) =0 .00283 log sin a=9. 95509 

logtan^a =9.79906 log tan \ a = 9 . 79906 colog sin 6= 0.00622 

log tan i (c + 6) = . 74265" log tan | (c - 6) = 7 . 99959 log sin ^4 = 9. 95921 

180° -i (c+6) = 79°44'53", H c ~ & ) = 0°34'21", 

\ (c + 6) = 100° 15' 7", J. and a in same quadrant § 66. 

c = 100° 49' 28", 6 =99° 40' 46", ^4 = 65° 33' 12". 

,«,..*• • * ■ ^ (log sin 5=9.99790 log sin 6=9 .99378 

Check, sin 5 sin c = sm 6 sin C } - gin c==9 . 99220 lo 5 sin C = 9. 99632 



9.99010 9.99010 



SOLUTION OF OBLIQUE SPHERICAL TRIANGLES 



85 



If only A is required, it can be found by natural functions from 
(120), or, cos A = —cos B cos C + sin B sin C cos a, or (44), cos A 
= — cos (B + C) — 2 sin B sin C sin 2 ^ a. 

69. Case III. Given a, b, A; solve the triangle. 

Assume first A and a in the same quadrant, a differing more 
from 90° than b. 

Construct the triangle, Fig. 29, making A equal to the given 
angle b equal to the given side, and with a as an arc radius describe 
a small circle BB' about C. It will in general intersect the side c 
in two points, B and B', and there will therefore be two solutions 




corresponding to the triangles ACB and ACB' when B and B' 
fall on the same side in front, and within 180° of A, since both 
contain all the given parts. 

The Ambiguous Case 

By the law of sines, sin A = sin B sin a/sin b, in which B may 
have either of its supplemental values, and the values of a, b, 
and A may be such as to give a solution for either. If a and A 
are in the same quadrant, and sin A is between zero and sin a/sin b, 
sin a being less than sin b, sin B will be less than unity and there 
will be two solutions. If sin A equals sin a/sin b, sin B will be 
unity, and two coincident right triangles will result. If sin A is 
greater than sin a/sin b, then sin B would be greater than unity 
and the solution is impossible. 

Hence, // a differs more from 90° than b, a and A will be in the 
same quadrant, and there will always be two solutions for values of 



sin A between and 



sm a 
sin b 



and none for other values. 



86 



SPHERICAL TRIGONOMETRY 



In the triangles, B will be in the same quadrant as A because the 
perpendicular from C to AB must be in the same quadrant with 
both, § 60. 

When a differs less from 90° than b, or b differs more from 90° 
than a, then b must be in the same quadrant with B, and therefore 
there can be but one solution. 

It should be noted that if the parts were allowed to vary between 
0° and 360° there would generally be two solutions. 

Example. Given a = 40° 16', b = 47° 44', A = 52° 30'; solve 
the triangle. 

sin A sin b 



From "Law of Sines," sin B 



Take either (136) or (137). 



sin a 
By first, 



(114*) 



tan J C = 



sin i (A + B) 



sin i(A - B) 
Take either (138) or (139). By first, 



tan J (a —6) 



cot \C 



sin \ (a + b) 



tan i(A - B) 



sin % (a — b) 

Since a differs more from 90 than b and is in the same quad- 
rant with A, there will be two solutions or none. 



log sin A = 9.89947 

log sin b = 9.86924 

colog sin a = 0.18953 

log sin B = 9.95824 

B = 65° 16' 30", 
B'= 114° 43' 30", 

log sin i (A + B) = 9.93255 

log tan i (6 - a) = 8.81459 

colog sin i (B - A) = 0.95369 

log tan i c = 9.70083 

^ c = 26 39 49, 

c = 53 19 38, 

log sin \ (a + 6) = 9.84177 

log tan J (£ - A) = 9.04901 

colog sin i (b - a) = 1.18633 

log cot \C = 0.07711 



J (A + 5) = 58° 53' 15". 
i (B - A) = 6° 23' 15". 
} (A + 5') = 83° 36' 45". 
i (£' - A) = 31° 6' 45". 
J (a - 6) = 44° 00' 00". 
i (b - a) = 3° 44' 00". 

log sin 



K4 + £') = 9.99729 



= 8.81459 
= 0.28674 
= 9.09862 



log tan J (b —a) 

colog sin J (B' — A) 

log tan ^ c' 

J c' = 7° 9' 10". 

c' = 14° 18' 20". 

log sin i (a .+ 6) = 9.84177 

log tan i (B f - A) = 9.78070 

colog sin i (b — a) = 1.18633 

log cot | C = 0.80880 



SOLUTION OF OBLIQUE SPHERICAL TRIANGLES 



87 



i C = 39° 56' 24", 

C = 79° 52' 48", 

5 = 65° 16' 30", 

c = 53° 19' 38", 



i C" = 8° 49' 41". 
C = 17° 39' 22". 
B = 114° 43' 30". 
c' = 14° 18' 20". 



Check, sin c sin j log sin c = 9.90421 log sin a = 9.81047 

A = sin a sin C \ log sin A = 9.89947 log sin C = 9.99320 

9.80368 9.80367 

C and c' may also be found from C and c. Draw a perpen- 
dicular CD, Fig. 26, to 55' forming a right triangle JSCZ), from 

cot B 



which tan * £C£' 



C"= C-2tan- 



(cos a) 
cot £ 



and tan | BB' = cos B tan a; whence 



cos 



— ) , and c' = c — 2 tan- 1 (cos B tan a), in 
a / 



which the algebraic sign of the last term must be observed. 




d~ b 



A' A 



Fig. 27. 




70. Case IV. Given A, B, a; solve the triangle. 

Assuming A to differ more from 90° than B, there will be two 
solutions. A and B may both be acute, both obtuse, or one 
obtuse and the other acute. In Fig. 27, construct a lune with 
the angle A. From the proportion sin A : sin B :: sin a : sin AC, 



find AC = sin- 1 



/ sin B sin a \ 
\ sin A ) ' 



and set it off from A to C, and 



with C as a center and a as an arc radius describe a small circle 
intersecting the other arc of the lune in B and B'. If B is acute 
the triangles of the two solutions will be ABC and A'B'C, but if 
B is obtuse, AB'C and A'BC, whether A and B are in the same 
or different quadrants, each set of triangles having all the parts. 



88 



SPHERICAL TRIGONOMETRY 



The Ambiguous Case 

By the same or similar suppositions on this as in Case III, 
these two laws are proved : 

// A differs more from 90° than B, A and a will be in the same 
quadrant, and there will be two solutions for all values of sin a 



between and 



sin 



sin B 



inclusive, and none for other values. 



If A differs less from 90° than B, or B differs more from 90° than 
A, B must be in the same quadrant as b, and therefore there can be 
but one solution. 

Example. Given A = 60°, B = 110°, a = 50°; solve the triangle. 

sin a sin B 



From "Law of Sines," sin b 



sin A 



(1144) 



Take either (136) or (137). By first, 



tan \c = 



sin \ {A 



B) 



tan i (a — b), 



sin \ (A - B) 
Take either (138) or (139). By first, 

coUC= sini^+|} tan (A _ 

sin i i/7. — M 



sin 



(a-b) 



log sin a 

log sin B 

colog sin A 



9.88425 
9.97299 
0.06247 



log sin b = 9.91971 

b = 56° 13' 22", 
V = 123° 46' 38", 



log sin i (A+ B) 



i (a + b) = 53° 6' 41" 
i(b - a) = 3° 6' 41" 
i ( a + b') = 86° 53' 19" 

i (V - a) = 36° 53' 19" 
i (A + B) = 85° 0' 0'. 
b(B - A) = 25° 0' 0'. 



9.99834 log sin J (A + B) = 9.99834 
log tan J (6 - a) = 8.73526 log tan J (6' - a) = 9.87536 
colog sin J (5 - A) = 0.37405 colog sin \ {B - A) = 0.37405 

log tan i c' = 0.24775 



log tan ^c = 9.10765 



1 r — 



7° 18' 5" ; 
c ^= 14° 36' 10", 



| c' = 60° 31' 20". 
c' = 121° 2' 40". 



L OF C. 



SOLUTION OF OBLIQUE SPHERICAL TRIANGLES 89 

log sin i (a + b) = 9.90298 log sin h (a + &') = 9.99936 

log tan i (B - A) = 9.66867 log tan } (B - A) = 9.66867 

c log sin \ (b - a) = 1.26538 colog sin J (&' - a) = 0.22166 

log cot J C = 0.83703 log cot J C = 9.88969 

§C= 8° 16' 50", JC" = 52° 11' 58". 

C = 16° 33' 40", (C = 104° 23' 56". 

6 = 56° 13' 21", \ b' = 123° 46' 38". 



c = 14° 36' 11", I C = 121° 2' 40". 

Check by the same formula and in the same way as in the 
preceding cases. As in the preceding case, C and c' may be 
found from C and c by a like determination of BCB' and £5' 

and substituting in C" = 180°- C + 2 tan - 1 /^2L? \ and c' = 180° 

\ cos a / 

— c + 2 tan -1 (cos B tan a), observing the signs of the last terms. 
71. Case V. Given a, b, c; solve the triangle. 

Any one of (125), (126), (127), may be used in this case, but 
when an angle is near 90° it cannot be accurately determined by 
its sine, nor one near by its cosine; it is therefore more accurate 
as well as shorter, when all the angles are required, to solve by 
(127): 

/ 
tan J A = y ' 

tan i B = 1/ 

tanJ-C=%/i 

V sins, sin (s — c) 

The work with these may be greatly shortened by assuming 



'sin (s — 


b) sin (s — c) 


sin s . 


sin (s — a) 


; sin (s — 


a) sin (s — c) 


sin s 


. sin (s — 6) 


sin (s — 


a) sin (s — 6) 



tan r = t/ S (s ~ a) sin ^ s ~ 6) sin (s Z c ) , (140) 
V sin s 



90 SPHERICAL TRIGONOMETRY 

which substituted above gives, 

UniA= , t / anr . tani5= tan r 



sin (s-a)' sin (s - 6) ' 

j. -> n tan r 

tan ^ C = . 

sin (s — c) 

Example. Given a = 114° 43' 18 r/ 7 6 = 136° 19' 36", c = 43° 
18' 30"; solve the triangle. 

s = 147° 10' 42", s - a = 32° 27' 24", s - 6 = 10° 51' 6", 
s - c = 103° 52' 12". 

colog sin s = 0.26598 log tan r = 9.62880 log tan r =9.62880 

log sin (s-a) = 9.72970 colog sin (s-a) = 0.27030 colog sin (s-b) = 0.72522 
log sin (s - b) = 9.27478 log tan i A = 9.89910 log tan § £ = 0.35402 

log sin (s-c)= 9.98714 £ A= 38° 24' 11" J 5= 66° 7' 39". 

2 log tan r= 9.25760 

log tan r = 9.62880 (A =76° 48' 22". 

colog sin | = Q01286 Ang _ 

log tan §C = 9.64166 

1(7=23° 39' 45". 



B = 132° 15' 18' 
C = 47° 19' 30" 



Check by the same formula as in preceding cases. 

72. Case VI. Given A, B, C; solve the triangle. 

The solution may be made by any one of (129), (130), (131) 
but more accurately, as in Case V, by (131). 

As in the preceding case, the computation in this may be sim- 
plified and shortened by assuming, 



tan R - y/ cog _ _ A) cog ™ & _ B) cog (S _ C) , (141) 

whence tan J a = tan R cos (S — A), tan J b = tan R cos 
OS - B), tan J c = tan R cos (S - C). 



SOLUTION OF OBLIQUE SPHERICAL TRIANGLES 91 

Example. Given A = 20° 9' 56", B = 55° 52' 32", C = 114° 
20' 14"; solve the triangle. 

5 = 95° 11' 21", 8 - A = 75° V 25", S - B = 39° 18' 49", 
5 - C = - 19° 8' 53". 



log (-cos S) =8.95638 log tan # = 9.84010 log tan # = 9.84010 

colog cos (S-A) =0.58768 log cos (S-A) =9.41232 log cos (S-B) =9.88857 
colog cos (S - B) = 0.1 1 143 log tan \ a = 9.25242 log tan \ b = 9.72867 

colog cos [s-C)= 0.02471 \ a = 10° 8' 19 1 " £6 = 28°9'50". 

2 log tan # = 9.68020 ( a = 2 0° 16' 38". 

log tan # = 9.84010 Ans - 6 = 56° 19' 40". 

log cos (S - C) = 9.97529 [ c = 66° 20' 48". 

log tan \c = 9.81539 

\ c = 33°10'24". 

Check the same as in other cases. 

These six cases might be solved by the methods of I, III, V, 
if II, IV* VI were reduced to their polar triangles, then solved 
and reduced back. But nothing would be gained in the way of 
saving labor. 



To Find the Area of a Spherical Triangle. 

73. In geometry it is proved that the area of a spherical tri- 
angle is its spherical excess E when the right angle is the unit of 
angles, and the trirectangular triangle is the unit of area. Hence 

+ i i, • i • +u • u+ i - + • A + B + C -180° 
the spherical excess in the right angle unit is . 

In radians, it is 

(A + B + O — n = 2 S - 7T, hence 

> 180° 

K = 2 S - 7i, also K = — tz } 

180° 



92 



SPHERICAL TRIGONOMETRY 



when the radius of the sphere is unity, § 56. If the radius is r, 
then since the surfaces of spheres are to each other as the squares 
of their radii, 

E 



K = r 2 (2 S - tc), or K = -=— nr 
" 180° 



(142) 



Exam-pie. Given A = 133° 12", B = 140° 18', C = 150° 30'; 
find the area of the triangle on a sphere of 5 miles radius. 



2 s = 133.2° + 140.3° + 150.5° 
180° 



7T = 1.3556 7T. 



K = 1.3556 it X 25 = 33.89 tt square miles. 

If the sides or sides and angles are given, find the angles by the 
corresponding case in the solutions of triangles, and substitute in 
(140). The solution with the given sides requires a difficult 
formula and more labor than the method suggested. 

74. To find the arc radius R of the circle circumscribing a spher- 
ical triangle. 




Fig. 28. 



In Fig. 28, O being the center of the required circle, let OAC 
x = OCA, and OA = R, then 

A = x + OAB, 

B = OB A + OBC = OCB + OAB, 
C = x + OCB. 
Adding, 
2 S = 2 x + 2 OC5 + 2 OA5 = 2z + 2£orz = £-£. 



-n i. x d tan 4 6 

But tan R = — 2 

cos (# — B) 



i, 



cos <S cos (S — i?) 
cos QSj^A) cos (S - C) 
cos (S - B) 



-v/ 



cos S 



cos OS - A) cos OS - B) cos (S-C) 



SOLUTION OF OBLIQUE SPHERICAL TRIANGLES 93 

which is (141), hence the auxiliary quantity assumed in the 
solution of Case VI, is the arc radius of the circumscribing circle 
of the triangle. 

75. To find the arc radius, r, oj the circle inscribed in a spherical 
triangle. 

Let the required radius in Fig. 29 be r and AD = x, AD = AE, 
DC = FC and EB = BF. Then, b = x + DC, c = x + EB, and 
a = FB + FC = EB + DC. Hence, 

2s = 2x + 2EB + 2DC = 2x + 2BF + 2FC = 2x + 2a, 

or s = x + a, .'. x = s — a. But sin x = sin (s — a) = tan r 
cot i A, 




or tan r = sin (s — a) tan J ^4. 



. , . /sin (s — b) sin (s — c) 

or = sin (s - a) t / ^ f--^ -^ 

y sin s. sin (s — a) 



tan r = \/ sin (s ~ a ) sin ^ s ~ ^ sin ( s ~ c ) • 
V sin s 

which is (142). 

Hence the auxiliary quantity assumed in the solution of Case V 
is the arc radius of the required inscribed circle. 



94 SPHERICAL TRIGONOMETRY 



EXAMPLES AND EXERCISES 

1. Given a = 120° 55' 35", b = 59° 4' 25", c = 106° 11' 22"; 

find A = 116° 44' 18", B = 63° 15' 42", C = 91° 8' 36". 

2. Given A = 128° 41' 49", B = 107° 33' 20", c = 124° 12' 31"; 

find a = 125° 41' 43", 'b = 82° 47' 33", C = 127° 22' 12". 

3. Given a = 88° 12' 20", 6 = 124° 7' 17", C = 50° 2' 1"; 

find A = 63° 15' 11", B = 132° 17' 59", c = 59° 4' 27". 

4. Given a = 150° 57' 5", 6 = 134° 15' 54", A = 144° 22' 42"; 

find 5 = 120° 36' 51", c = 55° 42' 8", C = 97° 43' 04", 
and B' = 59° 12' 15", c' = 23° 57' 18", C" = 29° 8' 46". 

5. Given A = 113° 39' 21", B = 123° 40' 18", a = 65° 39' 46"; 

find b = 124° 7' 20", c = 159° 50' 15", C = 159° 43' 34". 

6. Given A = 102° 14' 12", B = 54° 32' 24", C = 89° 5' 46"; 

find a = 104° 25' 9", b = 53° 49' 24", c = 97° 44' 19". 

7. Given ^ = 78° 19', B = 36° 15', c =112° 38'; 

find a = 101° 30' 59", b = 36° 16' 29", C = 112° 42' 48". 

8. Given a = 67° 35' 12", b = 58° 36' 6", A = 101° 17' 48"; 

find B = 64° 52' 50", c = 28° 1' 56", C = 29° 54' 5". 

9. Given A = 139° 54' 36", B = 61° 37' 54", a = 150° 17' 24"; 

find b = 42° 37' 22", c = 129° 40' 55", C = 89° 54' 19", 
and 6' = 137° 22' 38", c' = 19° 58' 40", C" = 26° 21' 23". 

10. Given a = 121° 17' 18", b = 76° 31' 18", C = 50° 12' 12"; 

find A = 133° 55' 52", B = 55° 2' 12", c = 65° 44' 26". 

11. Given a =112° 6' 42", b = 127° 39' 12", c = 71° 12' 42"; 

find A = 103° 52' 6", B = 123° 56' 10", C = 82° 47' 00". 

12. Given A = 120°, B = 130°, C = 80°; 

find c = 41° 44' 14", and also a and b. 

13. Given a = 30° 17' 36", b = 22° 14' 24", c = 18° 51' 48"; 

find A = 95° 30' 51", B = 48° 19' 09", C = 39° 55' 29". 

14. Given a = 156° 12' 12", b = 112° 48' 36", C = 76° 32' 24"; 

find A = 154° 4' 7", B = 87° 27' 03", c = 63° 48' 40". 

15. A = 140° 24' 36", B = 12° 18' 36", c = 28° 7' 42"; 

find a - 37° 58' 56", 6=11" 52' 54", C = 29° 13' 16". 



SOLUTION OF OBLIQUE SPHERICAL TRIANGLES 95 

16. Given A = 70°, B = 120°, b = 80; solve the triangle. 

17. Given A = 60°, B = 40°, b = 50° ; solve the triangle. 

18. Given a = 42°, B = 67°, b + c = 112°; (137), (35) 

find b = 52° 46' 33", A = 50° 40' 22", C = 83° 20' 30". 

19. Given a = 38°, B = 101°, A + C = 136°; (139), (35) 

find A = 41° 2' 48", b =113° 1' 40", c = 110° 55' 39". 

20. Given a = 48° 45' 40", b = 67° 12' 20", A = 42° 20' 30"; 

find 5 = 55° 40' 00", C = 116° 34' 09", c = 93° 8' 9", 
and £' = 124° 20' 00", C = 24° 32' 18", c' = 27° 37' 22". 

21. Given .4 = 36° 20' 20", B = 46° 30' 40", a = 42° 15' 20"; 

find b = 55° 25' 00", c = 81° 27' 14", C = 119° 22' 21", 
and 6' = 124° 35' 00", c' = C" = 

22. Given A = 60°, 5 = 110°, 6 =50°; find c = 11° 57' 48". 

23. Given A = 110°, 5 = 60°, 6 =50°; solve the triangle. 

24. Given a = 120° 30' 30", b = 69° 34' 56", A = 50° 10' 10"; solve 
the triangle. 

25. Given A = 95° 38' 4", (7 = 97° 26' 29", b = 64° 23' 15"; (§ 68) 

find B = 65° 33' 08". 

26. Given A' = 84° 20' 19", B = 27° 22' 40", c = 75° 33'; 

find E = 26159" and K = .12682 r 2 . 

27. Each side of an equilateral triangle on the earth's surface is 80° 
and the radius of the earth is 3960 miles. What is the area K. 

K = 17,647,000 square miles. 

28. Find the spherical excess of a triangle on the earth's surface whose 
sides are all 1°. Ans. 25". 

29. Find the angles between the faces of the five regular polyhedrons. 
SuggesUo,, *Qf *Qf - Q - -£}« IQZ 

108 

Ans. Tetrahedron = 70° 32', hexahedron = 90°, octahedron = 109° 28', 
dodecahedron = 116° 35', icosahedron = 138° 15'. 

30. A spherical square is divided into four equal right triangles by 
two diagonal arcs. Find the angle A of the square, having given the 
side 50°. Ans. 102° 33' 36". 



96 SPHERICAL TRIGONOMETRY 

31. Find the shortest distance between New York, lat. 40° 45' 24" N., 
long. 73° 58' 24" W., and Rio Janeiro, lat. 22° 54' 24" S., and long. 43° 
11' 24" W. 

Suggestions. New York to North Pole = 90° - 40° 45' 24", 
Rio Janeiro to North Pole = 00° + 22° 54' 24". 

Ans. 69° 47' 52". 



N.Pole 




Fig. 30. 

32. Find the distance between New York, lat. and long, as in 31, and 
Paris, lat. 48° 50' 12" N., and long. 2° 20' 12" E. 

Ans. 52° 26' 50". 

33. The area of an equilateral triangle is one fourth the area of the 
sphere. What are its angles and sides? 

Ans. 120°, 109° 28' 17". 

34. The latitude and longitude of one point on the earth are <f> and X, 
and that of another are <j> f and X', and d is the distance between them. 
Prove cos d — sin <j> sin <f> r + cos 4> cos <£' cos {X — X f ). 

35. If the angles of a spherical triangle be together equal to four 
right angles, prove by (130) and (78). Cos 2 \ a + cos 2 \ b + cos 2 £ c = 1. 

36. Prove sin * a S ' n * 6 = ^^S* . See (129), (130) 

sin \ c sin C 

or _ -r> sin \ a cos \ b cos (S — ^4) 

37. Prove —. — - — - — = ^ — - — - . 

sin \ c sin C 

38. Prove cos * a CQS * h = cos (S-Q # 

cos \ c sin C 

39. Prove tan fr B tan \ C = S1 " (* ■ ~ ^ . (127) 

sins 

40. Prove tan ^ 6 tan ^c = Z # g 4n • ( 131 > 

cos (o — A) 



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